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Concept of linear mappings are confusing me
Change of Basis ConfusionProve that a linear map for complex polynomials is diagonalizableEigenvalues of three given linear operatorsTransforming coordinate system vs objectsCan an $ntimes n$ matrix be reduced to a smaller matrix in any sense?Overview of Linear AlgebraLinear Transformation vs MatrixPruning SubsetsChange of basis formula - intuition/is this true?Linear 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
$endgroup$
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
$endgroup$
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
3 hours ago
$begingroup$
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
3 hours ago
$begingroup$
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
36 mins ago
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
$endgroup$
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
linear-algebra
asked 3 hours ago
mingming
4356
4356
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
3 hours ago
$begingroup$
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
3 hours ago
$begingroup$
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
36 mins ago
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
3 hours ago
$begingroup$
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
3 hours ago
$begingroup$
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
36 mins ago
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
3 hours ago
$begingroup$
maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
$endgroup$
– J. W. Tanner
3 hours ago
$begingroup$
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
3 hours ago
$begingroup$
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
3 hours ago
$begingroup$
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
36 mins ago
$begingroup$
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
36 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
32 mins ago
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 second coordinate?
Answer:
The matrix is $$left(beginarrayl1 & 0 & 0 \ 0 & 1 & 0 endarrayright)$$
Explanation:
To get the first row, think about what happens under matrix multiplication to the vector $(1,0,0)$. The next two rows 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.
$endgroup$
$begingroup$
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
34 mins ago
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.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
32 mins ago
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.
$endgroup$
$begingroup$
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
32 mins ago
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.
$endgroup$
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.
answered 3 hours ago
Ethan BolkerEthan Bolker
45.8k553120
45.8k553120
$begingroup$
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
32 mins ago
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
32 mins ago
$begingroup$
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
32 mins ago
$begingroup$
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
32 mins ago
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 second coordinate?
Answer:
The matrix is $$left(beginarrayl1 & 0 & 0 \ 0 & 1 & 0 endarrayright)$$
Explanation:
To get the first row, think about what happens under matrix multiplication to the vector $(1,0,0)$. The next two rows 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.
$endgroup$
$begingroup$
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
34 mins ago
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 second coordinate?
Answer:
The matrix is $$left(beginarrayl1 & 0 & 0 \ 0 & 1 & 0 endarrayright)$$
Explanation:
To get the first row, think about what happens under matrix multiplication to the vector $(1,0,0)$. The next two rows 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.
$endgroup$
$begingroup$
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
34 mins ago
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 second coordinate?
Answer:
The matrix is $$left(beginarrayl1 & 0 & 0 \ 0 & 1 & 0 endarrayright)$$
Explanation:
To get the first row, think about what happens under matrix multiplication to the vector $(1,0,0)$. The next two rows 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.
$endgroup$
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 second coordinate?
Answer:
The matrix is $$left(beginarrayl1 & 0 & 0 \ 0 & 1 & 0 endarrayright)$$
Explanation:
To get the first row, think about what happens under matrix multiplication to the vector $(1,0,0)$. The next two rows 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 1 hour ago
user156213user156213
64338
64338
$begingroup$
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
34 mins ago
add a comment |
$begingroup$
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
34 mins ago
$begingroup$
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
34 mins ago
$begingroup$
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
34 mins ago
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.
$endgroup$
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.
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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.
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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 2 hours ago
rschwiebrschwieb
108k12103253
108k12103253
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1
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maybe think of multiplying a matrix by a vector as a special case of multiplying a matrix by a matrix
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– J. W. Tanner
3 hours ago
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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 $mathbb R^2$?
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– John Douma
3 hours ago
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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
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– ming
36 mins ago