Mixing
Matrix multiplication, linear transformations and systems of equations
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In linear algebra, I learned that the reason why matrix multiplication is defined in its way is because of composition of linear transformations. So, multiplication of two matrices refers to the composition of two linear transformations.
But, when we look at a matrix representation of systems of equations, $$ Ax = b $$
where A is a m x n matrix, x is n x 1 matrix of n variables, and b is m x 1 matrix.
We apply matrix multiplication here, but we don't refer this to the composition of linear transformations since it is a representation of systems of equations.
I'm confused about why we do matrix multiplication in the way it's defined even though it does not refer to composition of linear transformations.
linear-algebra linear-transformations
asked Jan 16 at 18:45
JUNG WON CHOJUNG WON CHO
925
$endgroup$
add a comment |
$begingroup$
In linear algebra, I learned that the reason why matrix multiplication is defined in its way is because of composition of linear transformations. So, multiplication of two matrices refers to the composition of two linear transformations.
But, when we look at a matrix representation of systems of equations, $$ Ax = b $$
where A is a m x n matrix, x is n x 1 matrix of n variables, and b is m x 1 matrix.
We apply matrix multiplication here, but we don't refer this to the composition of linear transformations since it is a representation of systems of equations.
I'm confused about why we do matrix multiplication in the way it's defined even though it does not refer to composition of linear transformations.
linear-algebra linear-transformations
asked Jan 16 at 18:45
JUNG WON CHOJUNG WON CHO
925
$endgroup$
$begingroup$
If you're interested in developing your intuition on this topic, I would recommend the video series youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab. The third and fourth videos specifically should help out a bit.
$endgroup$
– Jack Pfaffinger
Jan 16 at 18:49
add a comment |
$begingroup$
In linear algebra, I learned that the reason why matrix multiplication is defined in its way is because of composition of linear transformations. So, multiplication of two matrices refers to the composition of two linear transformations.
But, when we look at a matrix representation of systems of equations, $$ Ax = b $$
where A is a m x n matrix, x is n x 1 matrix of n variables, and b is m x 1 matrix.
We apply matrix multiplication here, but we don't refer this to the composition of linear transformations since it is a representation of systems of equations.
I'm confused about why we do matrix multiplication in the way it's defined even though it does not refer to composition of linear transformations.
linear-algebra linear-transformations
asked Jan 16 at 18:45
JUNG WON CHOJUNG WON CHO
925
$endgroup$
In linear algebra, I learned that the reason why matrix multiplication is defined in its way is because of composition of linear transformations. So, multiplication of two matrices refers to the composition of two linear transformations.
But, when we look at a matrix representation of systems of equations, $$ Ax = b $$
where A is a m x n matrix, x is n x 1 matrix of n variables, and b is m x 1 matrix.
We apply matrix multiplication here, but we don't refer this to the composition of linear transformations since it is a representation of systems of equations.
I'm confused about why we do matrix multiplication in the way it's defined even though it does not refer to composition of linear transformations.
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Jan 16 at 18:45
JUNG WON CHOJUNG WON CHO
925
asked Jan 16 at 18:45
JUNG WON CHOJUNG WON CHO
925
asked Jan 16 at 18:45
JUNG WON CHOJUNG WON CHO
925
asked Jan 16 at 18:45
JUNG WON CHOJUNG WON CHO
925
asked Jan 16 at 18:45
JUNG WON CHOJUNG WON CHO
925
925
$begingroup$
If you're interested in developing your intuition on this topic, I would recommend the video series youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab. The third and fourth videos specifically should help out a bit.
$endgroup$
– Jack Pfaffinger
Jan 16 at 18:49
add a comment |
$begingroup$
If you're interested in developing your intuition on this topic, I would recommend the video series youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab. The third and fourth videos specifically should help out a bit.
$endgroup$
– Jack Pfaffinger
Jan 16 at 18:49
$begingroup$
If you're interested in developing your intuition on this topic, I would recommend the video series youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab. The third and fourth videos specifically should help out a bit.
$endgroup$
– Jack Pfaffinger
Jan 16 at 18:49
$begingroup$
If you're interested in developing your intuition on this topic, I would recommend the video series youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab. The third and fourth videos specifically should help out a bit.
$endgroup$
– Jack Pfaffinger
Jan 16 at 18:49
add a comment |
1 Answer
1
active
oldest
votes
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If $A$ is the matrix of a linear map $Lcolon Vlongrightarrow W$ with respect to two bases $B$ and $B'$ and if the coordinates of a vector $v$ with respect to the bases $B$ are $a_1,ldots,a_n$ (that is, if $v=a_1v_1+cdots+a_nv_n$, with $B={v_1,ldots,v_n}$), then the coordinates of $L(v)$ with respect to the basis $B'$ are precisely the entries of the vector$$B.begin{bmatrix}a_1\a_2\vdots\a_nend{bmatrix}$$So, multiplication of a matrix by a vector corresponds to computing the image of an element of $V$. And therefore, solving an equation $w=L(v)$ corresponds to a system of equations such as the one that you described.
answered Jan 16 at 19:05
José Carlos SantosJosé Carlos Santos
163k22131234
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add a comment |
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1 Answer
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active
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1 Answer
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oldest
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$begingroup$
If $A$ is the matrix of a linear map $Lcolon Vlongrightarrow W$ with respect to two bases $B$ and $B'$ and if the coordinates of a vector $v$ with respect to the bases $B$ are $a_1,ldots,a_n$ (that is, if $v=a_1v_1+cdots+a_nv_n$, with $B={v_1,ldots,v_n}$), then the coordinates of $L(v)$ with respect to the basis $B'$ are precisely the entries of the vector$$B.begin{bmatrix}a_1\a_2\vdots\a_nend{bmatrix}$$So, multiplication of a matrix by a vector corresponds to computing the image of an element of $V$. And therefore, solving an equation $w=L(v)$ corresponds to a system of equations such as the one that you described.
answered Jan 16 at 19:05
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
add a comment |
$begingroup$
If $A$ is the matrix of a linear map $Lcolon Vlongrightarrow W$ with respect to two bases $B$ and $B'$ and if the coordinates of a vector $v$ with respect to the bases $B$ are $a_1,ldots,a_n$ (that is, if $v=a_1v_1+cdots+a_nv_n$, with $B={v_1,ldots,v_n}$), then the coordinates of $L(v)$ with respect to the basis $B'$ are precisely the entries of the vector$$B.begin{bmatrix}a_1\a_2\vdots\a_nend{bmatrix}$$So, multiplication of a matrix by a vector corresponds to computing the image of an element of $V$. And therefore, solving an equation $w=L(v)$ corresponds to a system of equations such as the one that you described.
answered Jan 16 at 19:05
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
add a comment |
$begingroup$
If $A$ is the matrix of a linear map $Lcolon Vlongrightarrow W$ with respect to two bases $B$ and $B'$ and if the coordinates of a vector $v$ with respect to the bases $B$ are $a_1,ldots,a_n$ (that is, if $v=a_1v_1+cdots+a_nv_n$, with $B={v_1,ldots,v_n}$), then the coordinates of $L(v)$ with respect to the basis $B'$ are precisely the entries of the vector$$B.begin{bmatrix}a_1\a_2\vdots\a_nend{bmatrix}$$So, multiplication of a matrix by a vector corresponds to computing the image of an element of $V$. And therefore, solving an equation $w=L(v)$ corresponds to a system of equations such as the one that you described.
answered Jan 16 at 19:05
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
If $A$ is the matrix of a linear map $Lcolon Vlongrightarrow W$ with respect to two bases $B$ and $B'$ and if the coordinates of a vector $v$ with respect to the bases $B$ are $a_1,ldots,a_n$ (that is, if $v=a_1v_1+cdots+a_nv_n$, with $B={v_1,ldots,v_n}$), then the coordinates of $L(v)$ with respect to the basis $B'$ are precisely the entries of the vector$$B.begin{bmatrix}a_1\a_2\vdots\a_nend{bmatrix}$$So, multiplication of a matrix by a vector corresponds to computing the image of an element of $V$. And therefore, solving an equation $w=L(v)$ corresponds to a system of equations such as the one that you described.
answered Jan 16 at 19:05
José Carlos SantosJosé Carlos Santos
163k22131234
answered Jan 16 at 19:05
José Carlos SantosJosé Carlos Santos
163k22131234
answered Jan 16 at 19:05
José Carlos SantosJosé Carlos Santos
163k22131234
answered Jan 16 at 19:05
José Carlos SantosJosé Carlos Santos
163k22131234
163k22131234
add a comment |
add a comment |
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$begingroup$
If you're interested in developing your intuition on this topic, I would recommend the video series youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab. The third and fourth videos specifically should help out a bit.
$endgroup$
– Jack Pfaffinger
Jan 16 at 18:49