Why does a $2×3$ matrix multiplied by a vector in $BbbR^3$ give a vector in $BbbR^2$?Change of Basis ConfusionVariations in math to implement three-dimensional space?Transforming coordinate system vs objectsCan an $ntimes n$ matrix be reduced to a smaller matrix in any sense?What do groups and rings “look like”?Overview of Linear AlgebraZero subspace as solution spaceChange of basis formula - intuition/is this true?Expressing a matrix subspace as the image of a matrixLinear Algebra:Vector Space
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Why does a $2×3$ matrix multiplied by a vector in $BbbR^3$ give a vector in $BbbR^2$?
Change of Basis ConfusionVariations in math to implement three-dimensional space?Transforming coordinate system vs objectsCan an $ntimes n$ matrix be reduced to a smaller matrix in any sense?What do groups and rings “look like”?Overview of Linear AlgebraZero subspace as solution spaceChange of basis formula - intuition/is this true?Expressing a matrix subspace as the image of a matrixLinear Algebra:Vector Space
$begingroup$
I'm so confused on how we can have a 2x3 matrix A, multiply it by a vector in $Bbb R^3$ and then end up with a vector in $Bbb R^2$. Is it possible to visualize this at all or do I need to sort of blindly accept this concept as facts that I'll accept and use?
Can someone give a very brief summarization on why this makes sense? Because I just see it as, in a world (dimension) in $Bbb R^3$, we multiply it by a vector in $Bbb R^3$, and out pops a vector in $Bbb R^2$.
Thanks!
linear-algebra matrices vector-spaces vectors
edited Apr 8 at 6:28
user21820
40.5k544163
asked Apr 8 at 1:50
mingming
4606
$endgroup$
add a comment |
$begingroup$
I'm so confused on how we can have a 2x3 matrix A, multiply it by a vector in $Bbb R^3$ and then end up with a vector in $Bbb R^2$. Is it possible to visualize this at all or do I need to sort of blindly accept this concept as facts that I'll accept and use?
Can someone give a very brief summarization on why this makes sense? Because I just see it as, in a world (dimension) in $Bbb R^3$, we multiply it by a vector in $Bbb R^3$, and out pops a vector in $Bbb R^2$.
Thanks!
linear-algebra matrices vector-spaces vectors
edited Apr 8 at 6:28
user21820
40.5k544163
asked Apr 8 at 1:50
mingming
4606
$endgroup$
1
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
Apr 8 at 1:54
$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
Apr 8 at 2:06
$begingroup$
I think it's just visualizing it that gives me trouble. Like simple vector addition, I can easily say, oh ok just add the $x_1$ unit to the other $x_1$ unit and it stretches towards $x_1$'s side! But in this case, just multiplying a vector by something in one dimension and getting a vector in another dimensions just confuses me. And I do know we can have a function like that, it's just intuitively I guess I don't really understand it
$endgroup$
– ming
Apr 8 at 4:30
$begingroup$
If you think of vectors in $mathbb R^3$ as being three components in the $hat i$, $hat j$ and $hat k$ directions then you can think of the matrix as mapping $hat i$ to the first column vector, $hat j$ to the second column and $hat k$ to the third column. Since each of the column vectors are in two dimensions we end up with a vector in two dimensions.
$endgroup$
– John Douma
Apr 8 at 13:51
$begingroup$
Ohh.. ok thank you!
$endgroup$
– ming
Apr 8 at 22:56
add a comment |
$begingroup$
I'm so confused on how we can have a 2x3 matrix A, multiply it by a vector in $Bbb R^3$ and then end up with a vector in $Bbb R^2$. Is it possible to visualize this at all or do I need to sort of blindly accept this concept as facts that I'll accept and use?
Can someone give a very brief summarization on why this makes sense? Because I just see it as, in a world (dimension) in $Bbb R^3$, we multiply it by a vector in $Bbb R^3$, and out pops a vector in $Bbb R^2$.
Thanks!
linear-algebra matrices vector-spaces vectors
edited Apr 8 at 6:28
user21820
40.5k544163
asked Apr 8 at 1:50
mingming
4606
$endgroup$
I'm so confused on how we can have a 2x3 matrix A, multiply it by a vector in $Bbb R^3$ and then end up with a vector in $Bbb R^2$. Is it possible to visualize this at all or do I need to sort of blindly accept this concept as facts that I'll accept and use?
Can someone give a very brief summarization on why this makes sense? Because I just see it as, in a world (dimension) in $Bbb R^3$, we multiply it by a vector in $Bbb R^3$, and out pops a vector in $Bbb R^2$.
Thanks!
linear-algebra matrices vector-spaces vectors
linear-algebra matrices vector-spaces vectors
edited Apr 8 at 6:28
user21820
40.5k544163
asked Apr 8 at 1:50
mingming
4606
edited Apr 8 at 6:28
user21820
40.5k544163
asked Apr 8 at 1:50
mingming
4606
edited Apr 8 at 6:28
user21820
40.5k544163
edited Apr 8 at 6:28
user21820
40.5k544163
edited Apr 8 at 6:28
user21820
40.5k544163
40.5k544163
asked Apr 8 at 1:50
mingming
4606
asked Apr 8 at 1:50
mingming
4606
asked Apr 8 at 1:50
mingming
4606
4606
1
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
Apr 8 at 1:54
$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
Apr 8 at 2:06
$begingroup$
I think it's just visualizing it that gives me trouble. Like simple vector addition, I can easily say, oh ok just add the $x_1$ unit to the other $x_1$ unit and it stretches towards $x_1$'s side! But in this case, just multiplying a vector by something in one dimension and getting a vector in another dimensions just confuses me. And I do know we can have a function like that, it's just intuitively I guess I don't really understand it
$endgroup$
– ming
Apr 8 at 4:30
$begingroup$
If you think of vectors in $mathbb R^3$ as being three components in the $hat i$, $hat j$ and $hat k$ directions then you can think of the matrix as mapping $hat i$ to the first column vector, $hat j$ to the second column and $hat k$ to the third column. Since each of the column vectors are in two dimensions we end up with a vector in two dimensions.
$endgroup$
– John Douma
Apr 8 at 13:51
$begingroup$
Ohh.. ok thank you!
$endgroup$
– ming
Apr 8 at 22:56
add a comment |
1
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
Apr 8 at 1:54
$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
Apr 8 at 2:06
$begingroup$
I think it's just visualizing it that gives me trouble. Like simple vector addition, I can easily say, oh ok just add the $x_1$ unit to the other $x_1$ unit and it stretches towards $x_1$'s side! But in this case, just multiplying a vector by something in one dimension and getting a vector in another dimensions just confuses me. And I do know we can have a function like that, it's just intuitively I guess I don't really understand it
$endgroup$
– ming
Apr 8 at 4:30
$begingroup$
If you think of vectors in $mathbb R^3$ as being three components in the $hat i$, $hat j$ and $hat k$ directions then you can think of the matrix as mapping $hat i$ to the first column vector, $hat j$ to the second column and $hat k$ to the third column. Since each of the column vectors are in two dimensions we end up with a vector in two dimensions.
$endgroup$
– John Douma
Apr 8 at 13:51
$begingroup$
Ohh.. ok thank you!
$endgroup$
– ming
Apr 8 at 22:56
1
1
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
Apr 8 at 1:54
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
Apr 8 at 1:54
$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
Apr 8 at 2:06
$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
Apr 8 at 2:06
$begingroup$
I think it's just visualizing it that gives me trouble. Like simple vector addition, I can easily say, oh ok just add the $x_1$ unit to the other $x_1$ unit and it stretches towards $x_1$'s side! But in this case, just multiplying a vector by something in one dimension and getting a vector in another dimensions just confuses me. And I do know we can have a function like that, it's just intuitively I guess I don't really understand it
$endgroup$
– ming
Apr 8 at 4:30
$begingroup$
I think it's just visualizing it that gives me trouble. Like simple vector addition, I can easily say, oh ok just add the $x_1$ unit to the other $x_1$ unit and it stretches towards $x_1$'s side! But in this case, just multiplying a vector by something in one dimension and getting a vector in another dimensions just confuses me. And I do know we can have a function like that, it's just intuitively I guess I don't really understand it
$endgroup$
– ming
Apr 8 at 4:30
$begingroup$
If you think of vectors in $mathbb R^3$ as being three components in the $hat i$, $hat j$ and $hat k$ directions then you can think of the matrix as mapping $hat i$ to the first column vector, $hat j$ to the second column and $hat k$ to the third column. Since each of the column vectors are in two dimensions we end up with a vector in two dimensions.
$endgroup$
– John Douma
Apr 8 at 13:51
$begingroup$
If you think of vectors in $mathbb R^3$ as being three components in the $hat i$, $hat j$ and $hat k$ directions then you can think of the matrix as mapping $hat i$ to the first column vector, $hat j$ to the second column and $hat k$ to the third column. Since each of the column vectors are in two dimensions we end up with a vector in two dimensions.
$endgroup$
– John Douma
Apr 8 at 13:51
$begingroup$
Ohh.. ok thank you!
$endgroup$
– ming
Apr 8 at 22:56
$begingroup$
Ohh.. ok thank you!
$endgroup$
– ming
Apr 8 at 22:56
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
A more intuitive way is to think of a matrix "performing" on a vector, instead of a matrix "multiplying" with a vector.
Let's give an example. You have some triples of real numbers:
(1,2,3), (2,5,1), (3,5,9), (2,9,8)
and you "forget" the third coordinate:
(1,2), (2,5), (3,5), (2,9)
Surprisingly, this is an example of "matrix performance." Can you find
a matrix $M$ that "forgets" the third coordinate?
Answer:
The matrix is $$left(beginarrayl1 & 0 & 0 \ 0 & 1 & 0 endarrayright)$$
Explanation:
To get the first column, think about what happens under matrix multiplication to the vector $(1,0,0)$. The next two columns are similar.
We call such a matrix $M$ a projection.
We may visualize the projection as such.
Can you see what it means to "forget" the
third coordinate?
The important part of
a projection is linearity:
- You may project the addition of two vectors, or you may
add the projection of two vectors and you get the same result. - Similarly, you may project a scaled vector, or scale the vector
and then project it, and you get the same result.
We call a function with the linearity property a linear function.
In symbols, for any linear $f$,
- $f(v + w) = f(v) + f(w)$
- $f(cv) = cf(v)$
We see that the projection defined above is a
linear function.
Actually, you can check that every matrix is a linear function.
Perhaps it is more surprising that every linear function is a matrix. You may think of a matrix as a way to represent some linear function.
answered Apr 8 at 3:32
user156213user156213
69238
$endgroup$
2
$begingroup$
This was really helpful, thanks! Now if we "forget" about that third element, does that mean that the third dimension just totally disappears? The z-axis is just removed completely?
$endgroup$
– ming
Apr 8 at 4:32
$begingroup$
Yes, the third dimension "totally disappears!"
$endgroup$
– user156213
Apr 9 at 7:07
add a comment |
$begingroup$
For the moment don't think about multiplication and matrices.
You can imagine starting from a vector $(x,y,z)$ in $mathbbR^3$ and mapping it to a vector in $mathbbR^2$ this way, for example:
$$
(x, y, z) mapsto (2x+ z, 3x+ 4y).
$$
Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
$$
beginbmatrix
2 & 0 & 1 \
3 & 4 & 0
endbmatrix
beginbmatrix
1 \
2 \
3
endbmatrix
=
beginbmatrix
5\
11
endbmatrix.
$$
You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.
Edit in response to a comment.
No, this does not make $(5,11)$ "look like" $(1,2,3)$. Here is a toy example that suggests where you might find this kind of calculation. Suppose you run a business that builds three products. Call them A, B and C. To make an A you need $2$ widgets and $3$ gadgets. To make a B you need just $4$ gadgets. For a C you need just a single widget. How many widgets and gadgets should you order to make $1$ A, $2$ B's and $3$ C's? The matrix product above provides the answer. You could also use that $2 times 3$ matrix to figure out what orders you might fill if you knew how many widgets and gadgets you had in stock.
Matrices are helpful in geometry too. In a linear algebra course you learn how to see that when you use the matrix
$$
beginbmatrix
3 & -1 \
-1 & 3
endbmatrix
$$
to map the coordinate plane (pairs of numbers) to itself what you have done is stretch circles centered at the origin into ellipses by changing the scales along the diagonal lines $y=x$ and $y=-x$ m
answered Apr 8 at 2:06
Ethan BolkerEthan Bolker
47k555123
$endgroup$
$begingroup$
So in really simple terms, is (5, 11) a vector in 2 dimensions, that just "looks" like the vector (1, 2, 3) in 3 dimensions?
$endgroup$
– ming
Apr 8 at 4:33
$begingroup$
@ming No. See my edit.
$endgroup$
– Ethan Bolker
Apr 8 at 12:19
add a comment |
$begingroup$
A linear mapping has the property that it maps subspaces to subspaces.
So it will map a line to a line or $0$, a plane to a plane, a line, or $0$, and so on.
By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.
answered Apr 8 at 2:15
rschwiebrschwieb
108k12105253
$endgroup$
add a comment |
$begingroup$
Suppose you have a green tank, a blue tank, and a red tank, and suppose each liter in the blue tank contain .2 L of water and .1 L alcohol. For the blue tank, it's .3 L water .6 L alcohol. The red tank is .4 L water and .5 L alcohol. Now suppose we take $b$ liters from the blue tank, $g$ from the green, and $r$ from the red, and we look at how much total water ($w$) and alcohol ($a$) we have. We are starting with a three dimensional vector (how much from the green, blue, and red tanks), and ending up with a two dimensional vector (how much of water and alcohol). We can write this as:
$.2 g + .3 b + .4 r = w$
$.1 g + .6 b + .5 r = a$
In matrix form, that's
$$
beginbmatrix
.2 & .3 & .4 \
.1 & .6 & .5
endbmatrix
beginbmatrix
g \
b \
r
endbmatrix
=
beginbmatrix
w\
a
endbmatrix.
$$
Multiplying a vector be a matrix is simply a compact form of saying "take this much of each element"; in this case the $.2$ saying "take 20% of $b$ to get $w$", the $.1$ is saying "take 10% of $g$ to get $a$", etc. The column a number is in tells you which input number it's being multiplied with, and the row tells you what output it's contributing to.
answered Apr 8 at 15:11
AcccumulationAcccumulation
7,4612619
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A more intuitive way is to think of a matrix "performing" on a vector, instead of a matrix "multiplying" with a vector.
Let's give an example. You have some triples of real numbers:
(1,2,3), (2,5,1), (3,5,9), (2,9,8)
and you "forget" the third coordinate:
(1,2), (2,5), (3,5), (2,9)
Surprisingly, this is an example of "matrix performance." Can you find
a matrix $M$ that "forgets" the third coordinate?
Answer:
The matrix is $$left(beginarrayl1 & 0 & 0 \ 0 & 1 & 0 endarrayright)$$
Explanation:
To get the first column, think about what happens under matrix multiplication to the vector $(1,0,0)$. The next two columns are similar.
We call such a matrix $M$ a projection.
We may visualize the projection as such.
Can you see what it means to "forget" the
third coordinate?
The important part of
a projection is linearity:
- You may project the addition of two vectors, or you may
add the projection of two vectors and you get the same result. - Similarly, you may project a scaled vector, or scale the vector
and then project it, and you get the same result.
We call a function with the linearity property a linear function.
In symbols, for any linear $f$,
- $f(v + w) = f(v) + f(w)$
- $f(cv) = cf(v)$
We see that the projection defined above is a
linear function.
Actually, you can check that every matrix is a linear function.
Perhaps it is more surprising that every linear function is a matrix. You may think of a matrix as a way to represent some linear function.
answered Apr 8 at 3:32
user156213user156213
69238
$endgroup$
2
$begingroup$
This was really helpful, thanks! Now if we "forget" about that third element, does that mean that the third dimension just totally disappears? The z-axis is just removed completely?
$endgroup$
– ming
Apr 8 at 4:32
$begingroup$
Yes, the third dimension "totally disappears!"
$endgroup$
– user156213
Apr 9 at 7:07
add a comment |
$begingroup$
A more intuitive way is to think of a matrix "performing" on a vector, instead of a matrix "multiplying" with a vector.
Let's give an example. You have some triples of real numbers:
(1,2,3), (2,5,1), (3,5,9), (2,9,8)
and you "forget" the third coordinate:
(1,2), (2,5), (3,5), (2,9)
Surprisingly, this is an example of "matrix performance." Can you find
a matrix $M$ that "forgets" the third coordinate?
Answer:
The matrix is $$left(beginarrayl1 & 0 & 0 \ 0 & 1 & 0 endarrayright)$$
Explanation:
To get the first column, think about what happens under matrix multiplication to the vector $(1,0,0)$. The next two columns are similar.
We call such a matrix $M$ a projection.
We may visualize the projection as such.
Can you see what it means to "forget" the
third coordinate?
The important part of
a projection is linearity:
- You may project the addition of two vectors, or you may
add the projection of two vectors and you get the same result. - Similarly, you may project a scaled vector, or scale the vector
and then project it, and you get the same result.
We call a function with the linearity property a linear function.
In symbols, for any linear $f$,
- $f(v + w) = f(v) + f(w)$
- $f(cv) = cf(v)$
We see that the projection defined above is a
linear function.
Actually, you can check that every matrix is a linear function.
Perhaps it is more surprising that every linear function is a matrix. You may think of a matrix as a way to represent some linear function.
answered Apr 8 at 3:32
user156213user156213
69238
$endgroup$
2
$begingroup$
This was really helpful, thanks! Now if we "forget" about that third element, does that mean that the third dimension just totally disappears? The z-axis is just removed completely?
$endgroup$
– ming
Apr 8 at 4:32
$begingroup$
Yes, the third dimension "totally disappears!"
$endgroup$
– user156213
Apr 9 at 7:07
add a comment |
$begingroup$
A more intuitive way is to think of a matrix "performing" on a vector, instead of a matrix "multiplying" with a vector.
Let's give an example. You have some triples of real numbers:
(1,2,3), (2,5,1), (3,5,9), (2,9,8)
and you "forget" the third coordinate:
(1,2), (2,5), (3,5), (2,9)
Surprisingly, this is an example of "matrix performance." Can you find
a matrix $M$ that "forgets" the third coordinate?
Answer:
The matrix is $$left(beginarrayl1 & 0 & 0 \ 0 & 1 & 0 endarrayright)$$
Explanation:
To get the first column, think about what happens under matrix multiplication to the vector $(1,0,0)$. The next two columns are similar.
We call such a matrix $M$ a projection.
We may visualize the projection as such.
Can you see what it means to "forget" the
third coordinate?
The important part of
a projection is linearity:
- You may project the addition of two vectors, or you may
add the projection of two vectors and you get the same result. - Similarly, you may project a scaled vector, or scale the vector
and then project it, and you get the same result.
We call a function with the linearity property a linear function.
In symbols, for any linear $f$,
- $f(v + w) = f(v) + f(w)$
- $f(cv) = cf(v)$
We see that the projection defined above is a
linear function.
Actually, you can check that every matrix is a linear function.
Perhaps it is more surprising that every linear function is a matrix. You may think of a matrix as a way to represent some linear function.
answered Apr 8 at 3:32
user156213user156213
69238
$endgroup$
A more intuitive way is to think of a matrix "performing" on a vector, instead of a matrix "multiplying" with a vector.
Let's give an example. You have some triples of real numbers:
(1,2,3), (2,5,1), (3,5,9), (2,9,8)
and you "forget" the third coordinate:
(1,2), (2,5), (3,5), (2,9)
Surprisingly, this is an example of "matrix performance." Can you find
a matrix $M$ that "forgets" the third coordinate?
Answer:
The matrix is $$left(beginarrayl1 & 0 & 0 \ 0 & 1 & 0 endarrayright)$$
Explanation:
To get the first column, think about what happens under matrix multiplication to the vector $(1,0,0)$. The next two columns are similar.
We call such a matrix $M$ a projection.
We may visualize the projection as such.
Can you see what it means to "forget" the
third coordinate?
The important part of
a projection is linearity:
- You may project the addition of two vectors, or you may
add the projection of two vectors and you get the same result. - Similarly, you may project a scaled vector, or scale the vector
and then project it, and you get the same result.
We call a function with the linearity property a linear function.
In symbols, for any linear $f$,
- $f(v + w) = f(v) + f(w)$
- $f(cv) = cf(v)$
We see that the projection defined above is a
linear function.
Actually, you can check that every matrix is a linear function.
Perhaps it is more surprising that every linear function is a matrix. You may think of a matrix as a way to represent some linear function.
answered Apr 8 at 3:32
user156213user156213
69238
edited Apr 8 at 20:46
answered Apr 8 at 3:32
user156213user156213
69238
answered Apr 8 at 3:32
user156213user156213
69238
answered Apr 8 at 3:32
user156213user156213
69238
69238
2
$begingroup$
This was really helpful, thanks! Now if we "forget" about that third element, does that mean that the third dimension just totally disappears? The z-axis is just removed completely?
$endgroup$
– ming
Apr 8 at 4:32
$begingroup$
Yes, the third dimension "totally disappears!"
$endgroup$
– user156213
Apr 9 at 7:07
add a comment |
2
$begingroup$
This was really helpful, thanks! Now if we "forget" about that third element, does that mean that the third dimension just totally disappears? The z-axis is just removed completely?
$endgroup$
– ming
Apr 8 at 4:32
$begingroup$
Yes, the third dimension "totally disappears!"
$endgroup$
– user156213
Apr 9 at 7:07
2
2
$begingroup$
This was really helpful, thanks! Now if we "forget" about that third element, does that mean that the third dimension just totally disappears? The z-axis is just removed completely?
$endgroup$
– ming
Apr 8 at 4:32
$begingroup$
This was really helpful, thanks! Now if we "forget" about that third element, does that mean that the third dimension just totally disappears? The z-axis is just removed completely?
$endgroup$
– ming
Apr 8 at 4:32
$begingroup$
Yes, the third dimension "totally disappears!"
$endgroup$
– user156213
Apr 9 at 7:07
$begingroup$
Yes, the third dimension "totally disappears!"
$endgroup$
– user156213
Apr 9 at 7:07
add a comment |
$begingroup$
For the moment don't think about multiplication and matrices.
You can imagine starting from a vector $(x,y,z)$ in $mathbbR^3$ and mapping it to a vector in $mathbbR^2$ this way, for example:
$$
(x, y, z) mapsto (2x+ z, 3x+ 4y).
$$
Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
$$
beginbmatrix
2 & 0 & 1 \
3 & 4 & 0
endbmatrix
beginbmatrix
1 \
2 \
3
endbmatrix
=
beginbmatrix
5\
11
endbmatrix.
$$
You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.
Edit in response to a comment.
No, this does not make $(5,11)$ "look like" $(1,2,3)$. Here is a toy example that suggests where you might find this kind of calculation. Suppose you run a business that builds three products. Call them A, B and C. To make an A you need $2$ widgets and $3$ gadgets. To make a B you need just $4$ gadgets. For a C you need just a single widget. How many widgets and gadgets should you order to make $1$ A, $2$ B's and $3$ C's? The matrix product above provides the answer. You could also use that $2 times 3$ matrix to figure out what orders you might fill if you knew how many widgets and gadgets you had in stock.
Matrices are helpful in geometry too. In a linear algebra course you learn how to see that when you use the matrix
$$
beginbmatrix
3 & -1 \
-1 & 3
endbmatrix
$$
to map the coordinate plane (pairs of numbers) to itself what you have done is stretch circles centered at the origin into ellipses by changing the scales along the diagonal lines $y=x$ and $y=-x$ m
answered Apr 8 at 2:06
Ethan BolkerEthan Bolker
47k555123
$endgroup$
$begingroup$
So in really simple terms, is (5, 11) a vector in 2 dimensions, that just "looks" like the vector (1, 2, 3) in 3 dimensions?
$endgroup$
– ming
Apr 8 at 4:33
$begingroup$
@ming No. See my edit.
$endgroup$
– Ethan Bolker
Apr 8 at 12:19
add a comment |
$begingroup$
For the moment don't think about multiplication and matrices.
You can imagine starting from a vector $(x,y,z)$ in $mathbbR^3$ and mapping it to a vector in $mathbbR^2$ this way, for example:
$$
(x, y, z) mapsto (2x+ z, 3x+ 4y).
$$
Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
$$
beginbmatrix
2 & 0 & 1 \
3 & 4 & 0
endbmatrix
beginbmatrix
1 \
2 \
3
endbmatrix
=
beginbmatrix
5\
11
endbmatrix.
$$
You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.
Edit in response to a comment.
No, this does not make $(5,11)$ "look like" $(1,2,3)$. Here is a toy example that suggests where you might find this kind of calculation. Suppose you run a business that builds three products. Call them A, B and C. To make an A you need $2$ widgets and $3$ gadgets. To make a B you need just $4$ gadgets. For a C you need just a single widget. How many widgets and gadgets should you order to make $1$ A, $2$ B's and $3$ C's? The matrix product above provides the answer. You could also use that $2 times 3$ matrix to figure out what orders you might fill if you knew how many widgets and gadgets you had in stock.
Matrices are helpful in geometry too. In a linear algebra course you learn how to see that when you use the matrix
$$
beginbmatrix
3 & -1 \
-1 & 3
endbmatrix
$$
to map the coordinate plane (pairs of numbers) to itself what you have done is stretch circles centered at the origin into ellipses by changing the scales along the diagonal lines $y=x$ and $y=-x$ m
answered Apr 8 at 2:06
Ethan BolkerEthan Bolker
47k555123
$endgroup$
$begingroup$
So in really simple terms, is (5, 11) a vector in 2 dimensions, that just "looks" like the vector (1, 2, 3) in 3 dimensions?
$endgroup$
– ming
Apr 8 at 4:33
$begingroup$
@ming No. See my edit.
$endgroup$
– Ethan Bolker
Apr 8 at 12:19
add a comment |
$begingroup$
For the moment don't think about multiplication and matrices.
You can imagine starting from a vector $(x,y,z)$ in $mathbbR^3$ and mapping it to a vector in $mathbbR^2$ this way, for example:
$$
(x, y, z) mapsto (2x+ z, 3x+ 4y).
$$
Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
$$
beginbmatrix
2 & 0 & 1 \
3 & 4 & 0
endbmatrix
beginbmatrix
1 \
2 \
3
endbmatrix
=
beginbmatrix
5\
11
endbmatrix.
$$
You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.
Edit in response to a comment.
No, this does not make $(5,11)$ "look like" $(1,2,3)$. Here is a toy example that suggests where you might find this kind of calculation. Suppose you run a business that builds three products. Call them A, B and C. To make an A you need $2$ widgets and $3$ gadgets. To make a B you need just $4$ gadgets. For a C you need just a single widget. How many widgets and gadgets should you order to make $1$ A, $2$ B's and $3$ C's? The matrix product above provides the answer. You could also use that $2 times 3$ matrix to figure out what orders you might fill if you knew how many widgets and gadgets you had in stock.
Matrices are helpful in geometry too. In a linear algebra course you learn how to see that when you use the matrix
$$
beginbmatrix
3 & -1 \
-1 & 3
endbmatrix
$$
to map the coordinate plane (pairs of numbers) to itself what you have done is stretch circles centered at the origin into ellipses by changing the scales along the diagonal lines $y=x$ and $y=-x$ m
answered Apr 8 at 2:06
Ethan BolkerEthan Bolker
47k555123
$endgroup$
For the moment don't think about multiplication and matrices.
You can imagine starting from a vector $(x,y,z)$ in $mathbbR^3$ and mapping it to a vector in $mathbbR^2$ this way, for example:
$$
(x, y, z) mapsto (2x+ z, 3x+ 4y).
$$
Mathematicians have invented a nice clean way to write that map. It's the formalism you've learned for matrix multiplication. To see what $(1,2,3)$ maps to, calculate the matrix product
$$
beginbmatrix
2 & 0 & 1 \
3 & 4 & 0
endbmatrix
beginbmatrix
1 \
2 \
3
endbmatrix
=
beginbmatrix
5\
11
endbmatrix.
$$
You will soon be comfortable with this, just as you are now with whatever algorithm you were taught for ordinary multiplication. Then you will be free to focus on understanding what maps like this are useful for.
Edit in response to a comment.
No, this does not make $(5,11)$ "look like" $(1,2,3)$. Here is a toy example that suggests where you might find this kind of calculation. Suppose you run a business that builds three products. Call them A, B and C. To make an A you need $2$ widgets and $3$ gadgets. To make a B you need just $4$ gadgets. For a C you need just a single widget. How many widgets and gadgets should you order to make $1$ A, $2$ B's and $3$ C's? The matrix product above provides the answer. You could also use that $2 times 3$ matrix to figure out what orders you might fill if you knew how many widgets and gadgets you had in stock.
Matrices are helpful in geometry too. In a linear algebra course you learn how to see that when you use the matrix
$$
beginbmatrix
3 & -1 \
-1 & 3
endbmatrix
$$
to map the coordinate plane (pairs of numbers) to itself what you have done is stretch circles centered at the origin into ellipses by changing the scales along the diagonal lines $y=x$ and $y=-x$ m
answered Apr 8 at 2:06
Ethan BolkerEthan Bolker
47k555123
edited Apr 8 at 12:19
answered Apr 8 at 2:06
Ethan BolkerEthan Bolker
47k555123
answered Apr 8 at 2:06
Ethan BolkerEthan Bolker
47k555123
answered Apr 8 at 2:06
Ethan BolkerEthan Bolker
47k555123
47k555123
$begingroup$
So in really simple terms, is (5, 11) a vector in 2 dimensions, that just "looks" like the vector (1, 2, 3) in 3 dimensions?
$endgroup$
– ming
Apr 8 at 4:33
$begingroup$
@ming No. See my edit.
$endgroup$
– Ethan Bolker
Apr 8 at 12:19
add a comment |
$begingroup$
So in really simple terms, is (5, 11) a vector in 2 dimensions, that just "looks" like the vector (1, 2, 3) in 3 dimensions?
$endgroup$
– ming
Apr 8 at 4:33
$begingroup$
@ming No. See my edit.
$endgroup$
– Ethan Bolker
Apr 8 at 12:19
$begingroup$
So in really simple terms, is (5, 11) a vector in 2 dimensions, that just "looks" like the vector (1, 2, 3) in 3 dimensions?
$endgroup$
– ming
Apr 8 at 4:33
$begingroup$
So in really simple terms, is (5, 11) a vector in 2 dimensions, that just "looks" like the vector (1, 2, 3) in 3 dimensions?
$endgroup$
– ming
Apr 8 at 4:33
$begingroup$
@ming No. See my edit.
$endgroup$
– Ethan Bolker
Apr 8 at 12:19
$begingroup$
@ming No. See my edit.
$endgroup$
– Ethan Bolker
Apr 8 at 12:19
add a comment |
$begingroup$
A linear mapping has the property that it maps subspaces to subspaces.
So it will map a line to a line or $0$, a plane to a plane, a line, or $0$, and so on.
By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.
answered Apr 8 at 2:15
rschwiebrschwieb
108k12105253
$endgroup$
add a comment |
$begingroup$
A linear mapping has the property that it maps subspaces to subspaces.
So it will map a line to a line or $0$, a plane to a plane, a line, or $0$, and so on.
By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.
answered Apr 8 at 2:15
rschwiebrschwieb
108k12105253
$endgroup$
add a comment |
$begingroup$
A linear mapping has the property that it maps subspaces to subspaces.
So it will map a line to a line or $0$, a plane to a plane, a line, or $0$, and so on.
By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.
answered Apr 8 at 2:15
rschwiebrschwieb
108k12105253
$endgroup$
A linear mapping has the property that it maps subspaces to subspaces.
So it will map a line to a line or $0$, a plane to a plane, a line, or $0$, and so on.
By definition, linear mappings “play nice” with addition and scaling. These properties allow us to reduce statements about entire vector spaces down to bases, which are quite “small” in the finite dimensional case.
answered Apr 8 at 2:15
rschwiebrschwieb
108k12105253
answered Apr 8 at 2:15
rschwiebrschwieb
108k12105253
answered Apr 8 at 2:15
rschwiebrschwieb
108k12105253
answered Apr 8 at 2:15
rschwiebrschwieb
108k12105253
108k12105253
add a comment |
add a comment |
$begingroup$
Suppose you have a green tank, a blue tank, and a red tank, and suppose each liter in the blue tank contain .2 L of water and .1 L alcohol. For the blue tank, it's .3 L water .6 L alcohol. The red tank is .4 L water and .5 L alcohol. Now suppose we take $b$ liters from the blue tank, $g$ from the green, and $r$ from the red, and we look at how much total water ($w$) and alcohol ($a$) we have. We are starting with a three dimensional vector (how much from the green, blue, and red tanks), and ending up with a two dimensional vector (how much of water and alcohol). We can write this as:
$.2 g + .3 b + .4 r = w$
$.1 g + .6 b + .5 r = a$
In matrix form, that's
$$
beginbmatrix
.2 & .3 & .4 \
.1 & .6 & .5
endbmatrix
beginbmatrix
g \
b \
r
endbmatrix
=
beginbmatrix
w\
a
endbmatrix.
$$
Multiplying a vector be a matrix is simply a compact form of saying "take this much of each element"; in this case the $.2$ saying "take 20% of $b$ to get $w$", the $.1$ is saying "take 10% of $g$ to get $a$", etc. The column a number is in tells you which input number it's being multiplied with, and the row tells you what output it's contributing to.
answered Apr 8 at 15:11
AcccumulationAcccumulation
7,4612619
$endgroup$
add a comment |
$begingroup$
Suppose you have a green tank, a blue tank, and a red tank, and suppose each liter in the blue tank contain .2 L of water and .1 L alcohol. For the blue tank, it's .3 L water .6 L alcohol. The red tank is .4 L water and .5 L alcohol. Now suppose we take $b$ liters from the blue tank, $g$ from the green, and $r$ from the red, and we look at how much total water ($w$) and alcohol ($a$) we have. We are starting with a three dimensional vector (how much from the green, blue, and red tanks), and ending up with a two dimensional vector (how much of water and alcohol). We can write this as:
$.2 g + .3 b + .4 r = w$
$.1 g + .6 b + .5 r = a$
In matrix form, that's
$$
beginbmatrix
.2 & .3 & .4 \
.1 & .6 & .5
endbmatrix
beginbmatrix
g \
b \
r
endbmatrix
=
beginbmatrix
w\
a
endbmatrix.
$$
Multiplying a vector be a matrix is simply a compact form of saying "take this much of each element"; in this case the $.2$ saying "take 20% of $b$ to get $w$", the $.1$ is saying "take 10% of $g$ to get $a$", etc. The column a number is in tells you which input number it's being multiplied with, and the row tells you what output it's contributing to.
answered Apr 8 at 15:11
AcccumulationAcccumulation
7,4612619
$endgroup$
add a comment |
$begingroup$
Suppose you have a green tank, a blue tank, and a red tank, and suppose each liter in the blue tank contain .2 L of water and .1 L alcohol. For the blue tank, it's .3 L water .6 L alcohol. The red tank is .4 L water and .5 L alcohol. Now suppose we take $b$ liters from the blue tank, $g$ from the green, and $r$ from the red, and we look at how much total water ($w$) and alcohol ($a$) we have. We are starting with a three dimensional vector (how much from the green, blue, and red tanks), and ending up with a two dimensional vector (how much of water and alcohol). We can write this as:
$.2 g + .3 b + .4 r = w$
$.1 g + .6 b + .5 r = a$
In matrix form, that's
$$
beginbmatrix
.2 & .3 & .4 \
.1 & .6 & .5
endbmatrix
beginbmatrix
g \
b \
r
endbmatrix
=
beginbmatrix
w\
a
endbmatrix.
$$
Multiplying a vector be a matrix is simply a compact form of saying "take this much of each element"; in this case the $.2$ saying "take 20% of $b$ to get $w$", the $.1$ is saying "take 10% of $g$ to get $a$", etc. The column a number is in tells you which input number it's being multiplied with, and the row tells you what output it's contributing to.
answered Apr 8 at 15:11
AcccumulationAcccumulation
7,4612619
$endgroup$
Suppose you have a green tank, a blue tank, and a red tank, and suppose each liter in the blue tank contain .2 L of water and .1 L alcohol. For the blue tank, it's .3 L water .6 L alcohol. The red tank is .4 L water and .5 L alcohol. Now suppose we take $b$ liters from the blue tank, $g$ from the green, and $r$ from the red, and we look at how much total water ($w$) and alcohol ($a$) we have. We are starting with a three dimensional vector (how much from the green, blue, and red tanks), and ending up with a two dimensional vector (how much of water and alcohol). We can write this as:
$.2 g + .3 b + .4 r = w$
$.1 g + .6 b + .5 r = a$
In matrix form, that's
$$
beginbmatrix
.2 & .3 & .4 \
.1 & .6 & .5
endbmatrix
beginbmatrix
g \
b \
r
endbmatrix
=
beginbmatrix
w\
a
endbmatrix.
$$
Multiplying a vector be a matrix is simply a compact form of saying "take this much of each element"; in this case the $.2$ saying "take 20% of $b$ to get $w$", the $.1$ is saying "take 10% of $g$ to get $a$", etc. The column a number is in tells you which input number it's being multiplied with, and the row tells you what output it's contributing to.
answered Apr 8 at 15:11
AcccumulationAcccumulation
7,4612619
answered Apr 8 at 15:11
AcccumulationAcccumulation
7,4612619
answered Apr 8 at 15:11
AcccumulationAcccumulation
7,4612619
answered Apr 8 at 15:11
AcccumulationAcccumulation
7,4612619
7,4612619
add a comment |
add a comment |
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1
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
Apr 8 at 1:54
$begingroup$
Is it the definition of matrix multiplication that gives you trouble? Have you tried doing a multiplication and seeing what you get? Do you understand that we can have a function like $f(x,y,z)=(x-2y+z, 2x+4y-z)$ which maps $mathbb R^3$ to $mathbb R^2$?
$endgroup$
– John Douma
Apr 8 at 2:06
$begingroup$
I think it's just visualizing it that gives me trouble. Like simple vector addition, I can easily say, oh ok just add the $x_1$ unit to the other $x_1$ unit and it stretches towards $x_1$'s side! But in this case, just multiplying a vector by something in one dimension and getting a vector in another dimensions just confuses me. And I do know we can have a function like that, it's just intuitively I guess I don't really understand it
$endgroup$
– ming
Apr 8 at 4:30
$begingroup$
If you think of vectors in $mathbb R^3$ as being three components in the $hat i$, $hat j$ and $hat k$ directions then you can think of the matrix as mapping $hat i$ to the first column vector, $hat j$ to the second column and $hat k$ to the third column. Since each of the column vectors are in two dimensions we end up with a vector in two dimensions.
$endgroup$
– John Douma
Apr 8 at 13:51
$begingroup$
Ohh.. ok thank you!
$endgroup$
– ming
Apr 8 at 22:56