Mixing
Reconciling the 'column' and the 'row' pictures for matrices
$begingroup$
I understand the column picture (namely considering a matrix equation $Ax=b$ as a vector equation with each column of the matrix $A$ scalarly multiplied by each coefficient of the vector $x$ to give the vector $b$) and the row picture (intersection of the planes, or hyperplanes/lines according to the dimensions).
But since these are equivalent, there has to be some way to show this equivalency? What I mean is, the vectors in the column picture must somehow directly correspond to the planes in the row picture, and the procedures we follow in either of the pictures must be equivalent. Is there an easy way to visualise this equivalency?
I do not mean equivalence, in terms of determining the solutions, that is, to determine if the planes are parallel based on the column vectors. What I mean is to get to a system/visualisation, which can connect the two pictures.
The row and the column pictures are as mentioned in the Introduction to Linear Algebra by Gilbert Strang.
linear-algebra matrices systems-of-equations matrix-equations
asked Jul 6 '17 at 10:16
Satwik PasaniSatwik Pasani
313317
$endgroup$
add a comment |
$begingroup$
I understand the column picture (namely considering a matrix equation $Ax=b$ as a vector equation with each column of the matrix $A$ scalarly multiplied by each coefficient of the vector $x$ to give the vector $b$) and the row picture (intersection of the planes, or hyperplanes/lines according to the dimensions).
But since these are equivalent, there has to be some way to show this equivalency? What I mean is, the vectors in the column picture must somehow directly correspond to the planes in the row picture, and the procedures we follow in either of the pictures must be equivalent. Is there an easy way to visualise this equivalency?
I do not mean equivalence, in terms of determining the solutions, that is, to determine if the planes are parallel based on the column vectors. What I mean is to get to a system/visualisation, which can connect the two pictures.
The row and the column pictures are as mentioned in the Introduction to Linear Algebra by Gilbert Strang.
linear-algebra matrices systems-of-equations matrix-equations
asked Jul 6 '17 at 10:16
Satwik PasaniSatwik Pasani
313317
$endgroup$
$begingroup$
Linked with the topic web.mit.edu/18.06/www/Essays/newpaper_ver3.pdf
$endgroup$
– Widawensen
Jul 6 '17 at 10:29
add a comment |
$begingroup$
I understand the column picture (namely considering a matrix equation $Ax=b$ as a vector equation with each column of the matrix $A$ scalarly multiplied by each coefficient of the vector $x$ to give the vector $b$) and the row picture (intersection of the planes, or hyperplanes/lines according to the dimensions).
But since these are equivalent, there has to be some way to show this equivalency? What I mean is, the vectors in the column picture must somehow directly correspond to the planes in the row picture, and the procedures we follow in either of the pictures must be equivalent. Is there an easy way to visualise this equivalency?
I do not mean equivalence, in terms of determining the solutions, that is, to determine if the planes are parallel based on the column vectors. What I mean is to get to a system/visualisation, which can connect the two pictures.
The row and the column pictures are as mentioned in the Introduction to Linear Algebra by Gilbert Strang.
linear-algebra matrices systems-of-equations matrix-equations
asked Jul 6 '17 at 10:16
Satwik PasaniSatwik Pasani
313317
$endgroup$
I understand the column picture (namely considering a matrix equation $Ax=b$ as a vector equation with each column of the matrix $A$ scalarly multiplied by each coefficient of the vector $x$ to give the vector $b$) and the row picture (intersection of the planes, or hyperplanes/lines according to the dimensions).
But since these are equivalent, there has to be some way to show this equivalency? What I mean is, the vectors in the column picture must somehow directly correspond to the planes in the row picture, and the procedures we follow in either of the pictures must be equivalent. Is there an easy way to visualise this equivalency?
I do not mean equivalence, in terms of determining the solutions, that is, to determine if the planes are parallel based on the column vectors. What I mean is to get to a system/visualisation, which can connect the two pictures.
The row and the column pictures are as mentioned in the Introduction to Linear Algebra by Gilbert Strang.
linear-algebra matrices systems-of-equations matrix-equations
linear-algebra matrices systems-of-equations matrix-equations
asked Jul 6 '17 at 10:16
Satwik PasaniSatwik Pasani
313317
asked Jul 6 '17 at 10:16
Satwik PasaniSatwik Pasani
313317
asked Jul 6 '17 at 10:16
Satwik PasaniSatwik Pasani
313317
asked Jul 6 '17 at 10:16
Satwik PasaniSatwik Pasani
313317
asked Jul 6 '17 at 10:16
Satwik PasaniSatwik Pasani
313317
313317
$begingroup$
Linked with the topic web.mit.edu/18.06/www/Essays/newpaper_ver3.pdf
$endgroup$
– Widawensen
Jul 6 '17 at 10:29
add a comment |
$begingroup$
Linked with the topic web.mit.edu/18.06/www/Essays/newpaper_ver3.pdf
$endgroup$
– Widawensen
Jul 6 '17 at 10:29
$begingroup$
Linked with the topic web.mit.edu/18.06/www/Essays/newpaper_ver3.pdf
$endgroup$
– Widawensen
Jul 6 '17 at 10:29
$begingroup$
Linked with the topic web.mit.edu/18.06/www/Essays/newpaper_ver3.pdf
$endgroup$
– Widawensen
Jul 6 '17 at 10:29
add a comment |
1 Answer
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$begingroup$
You must realise that a $ntimes m$ matrix $A$ encodes a linear map $Bbb R^mtoBbb R^n$ (note the order of $m,n$) and that the spaces at departure and arrival are otherwise unrelated. Moreover, matrices are usually just a tool to numerically represent a linear map $f:Vto W$ between two more abstract vector spaces, after choosing a basis in both; that actual matrix entries one gets depend heavily on the basis used. As a consequence there cannot be a direct geometrical relation between columns, which represent vectors of $W$ in coordinates with respect to the basis chosen there, and rows, which represent linear combinations of coordinate functions (for the chosen basis) on$~V$.
Nonetheless there is a way to understand what is going on from two points of view. Solving a system of equation means trying to find the pre-image$~x$ (which must live in$~V$) under $f$ of a given vector $yin W$. The column picture says that the columns of $A$ describe particular vectors in$~W$ that are $f$-images of the chosen basis in$~V$; the question then is to find a linear combination of these that equals the given$~y$, and the corresponding linear combination of the basis vectors of $V$ will be our solution $x$ (of course there might be no solution, or infinitely many of them). The row picture uses the coordinates in$~W$ with respect to the chosen basis, one at a time. In order for an $f$-image to match$~y$, each of its coordinates must be right. Computing one coordinate of the $f$-image of an unknown $x$ amounts to computing a linear combination of the coordinates of$~x$ with respect to the basis chosen in $V$, and the coefficients of such a linear combination are given by a row of $A$. Requiring that one coordinates of the image matches the one of $y$ means that this linear combination of the coordinates of$~x$ must have a specific value, and that (usually) defines a plane in $V$. Combining these requirements means intersecting the planes in $V$ to find te set of possibilities for$~x$.
answered Jul 6 '17 at 10:28
Marc van LeeuwenMarc van Leeuwen
87.9k5111226
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add a comment |
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$begingroup$
You must realise that a $ntimes m$ matrix $A$ encodes a linear map $Bbb R^mtoBbb R^n$ (note the order of $m,n$) and that the spaces at departure and arrival are otherwise unrelated. Moreover, matrices are usually just a tool to numerically represent a linear map $f:Vto W$ between two more abstract vector spaces, after choosing a basis in both; that actual matrix entries one gets depend heavily on the basis used. As a consequence there cannot be a direct geometrical relation between columns, which represent vectors of $W$ in coordinates with respect to the basis chosen there, and rows, which represent linear combinations of coordinate functions (for the chosen basis) on$~V$.
Nonetheless there is a way to understand what is going on from two points of view. Solving a system of equation means trying to find the pre-image$~x$ (which must live in$~V$) under $f$ of a given vector $yin W$. The column picture says that the columns of $A$ describe particular vectors in$~W$ that are $f$-images of the chosen basis in$~V$; the question then is to find a linear combination of these that equals the given$~y$, and the corresponding linear combination of the basis vectors of $V$ will be our solution $x$ (of course there might be no solution, or infinitely many of them). The row picture uses the coordinates in$~W$ with respect to the chosen basis, one at a time. In order for an $f$-image to match$~y$, each of its coordinates must be right. Computing one coordinate of the $f$-image of an unknown $x$ amounts to computing a linear combination of the coordinates of$~x$ with respect to the basis chosen in $V$, and the coefficients of such a linear combination are given by a row of $A$. Requiring that one coordinates of the image matches the one of $y$ means that this linear combination of the coordinates of$~x$ must have a specific value, and that (usually) defines a plane in $V$. Combining these requirements means intersecting the planes in $V$ to find te set of possibilities for$~x$.
answered Jul 6 '17 at 10:28
Marc van LeeuwenMarc van Leeuwen
87.9k5111226
$endgroup$
add a comment |
$begingroup$
You must realise that a $ntimes m$ matrix $A$ encodes a linear map $Bbb R^mtoBbb R^n$ (note the order of $m,n$) and that the spaces at departure and arrival are otherwise unrelated. Moreover, matrices are usually just a tool to numerically represent a linear map $f:Vto W$ between two more abstract vector spaces, after choosing a basis in both; that actual matrix entries one gets depend heavily on the basis used. As a consequence there cannot be a direct geometrical relation between columns, which represent vectors of $W$ in coordinates with respect to the basis chosen there, and rows, which represent linear combinations of coordinate functions (for the chosen basis) on$~V$.
Nonetheless there is a way to understand what is going on from two points of view. Solving a system of equation means trying to find the pre-image$~x$ (which must live in$~V$) under $f$ of a given vector $yin W$. The column picture says that the columns of $A$ describe particular vectors in$~W$ that are $f$-images of the chosen basis in$~V$; the question then is to find a linear combination of these that equals the given$~y$, and the corresponding linear combination of the basis vectors of $V$ will be our solution $x$ (of course there might be no solution, or infinitely many of them). The row picture uses the coordinates in$~W$ with respect to the chosen basis, one at a time. In order for an $f$-image to match$~y$, each of its coordinates must be right. Computing one coordinate of the $f$-image of an unknown $x$ amounts to computing a linear combination of the coordinates of$~x$ with respect to the basis chosen in $V$, and the coefficients of such a linear combination are given by a row of $A$. Requiring that one coordinates of the image matches the one of $y$ means that this linear combination of the coordinates of$~x$ must have a specific value, and that (usually) defines a plane in $V$. Combining these requirements means intersecting the planes in $V$ to find te set of possibilities for$~x$.
answered Jul 6 '17 at 10:28
Marc van LeeuwenMarc van Leeuwen
87.9k5111226
$endgroup$
add a comment |
$begingroup$
You must realise that a $ntimes m$ matrix $A$ encodes a linear map $Bbb R^mtoBbb R^n$ (note the order of $m,n$) and that the spaces at departure and arrival are otherwise unrelated. Moreover, matrices are usually just a tool to numerically represent a linear map $f:Vto W$ between two more abstract vector spaces, after choosing a basis in both; that actual matrix entries one gets depend heavily on the basis used. As a consequence there cannot be a direct geometrical relation between columns, which represent vectors of $W$ in coordinates with respect to the basis chosen there, and rows, which represent linear combinations of coordinate functions (for the chosen basis) on$~V$.
Nonetheless there is a way to understand what is going on from two points of view. Solving a system of equation means trying to find the pre-image$~x$ (which must live in$~V$) under $f$ of a given vector $yin W$. The column picture says that the columns of $A$ describe particular vectors in$~W$ that are $f$-images of the chosen basis in$~V$; the question then is to find a linear combination of these that equals the given$~y$, and the corresponding linear combination of the basis vectors of $V$ will be our solution $x$ (of course there might be no solution, or infinitely many of them). The row picture uses the coordinates in$~W$ with respect to the chosen basis, one at a time. In order for an $f$-image to match$~y$, each of its coordinates must be right. Computing one coordinate of the $f$-image of an unknown $x$ amounts to computing a linear combination of the coordinates of$~x$ with respect to the basis chosen in $V$, and the coefficients of such a linear combination are given by a row of $A$. Requiring that one coordinates of the image matches the one of $y$ means that this linear combination of the coordinates of$~x$ must have a specific value, and that (usually) defines a plane in $V$. Combining these requirements means intersecting the planes in $V$ to find te set of possibilities for$~x$.
answered Jul 6 '17 at 10:28
Marc van LeeuwenMarc van Leeuwen
87.9k5111226
$endgroup$
You must realise that a $ntimes m$ matrix $A$ encodes a linear map $Bbb R^mtoBbb R^n$ (note the order of $m,n$) and that the spaces at departure and arrival are otherwise unrelated. Moreover, matrices are usually just a tool to numerically represent a linear map $f:Vto W$ between two more abstract vector spaces, after choosing a basis in both; that actual matrix entries one gets depend heavily on the basis used. As a consequence there cannot be a direct geometrical relation between columns, which represent vectors of $W$ in coordinates with respect to the basis chosen there, and rows, which represent linear combinations of coordinate functions (for the chosen basis) on$~V$.
Nonetheless there is a way to understand what is going on from two points of view. Solving a system of equation means trying to find the pre-image$~x$ (which must live in$~V$) under $f$ of a given vector $yin W$. The column picture says that the columns of $A$ describe particular vectors in$~W$ that are $f$-images of the chosen basis in$~V$; the question then is to find a linear combination of these that equals the given$~y$, and the corresponding linear combination of the basis vectors of $V$ will be our solution $x$ (of course there might be no solution, or infinitely many of them). The row picture uses the coordinates in$~W$ with respect to the chosen basis, one at a time. In order for an $f$-image to match$~y$, each of its coordinates must be right. Computing one coordinate of the $f$-image of an unknown $x$ amounts to computing a linear combination of the coordinates of$~x$ with respect to the basis chosen in $V$, and the coefficients of such a linear combination are given by a row of $A$. Requiring that one coordinates of the image matches the one of $y$ means that this linear combination of the coordinates of$~x$ must have a specific value, and that (usually) defines a plane in $V$. Combining these requirements means intersecting the planes in $V$ to find te set of possibilities for$~x$.
answered Jul 6 '17 at 10:28
Marc van LeeuwenMarc van Leeuwen
87.9k5111226
edited Jan 21 at 12:06
answered Jul 6 '17 at 10:28
Marc van LeeuwenMarc van Leeuwen
87.9k5111226
answered Jul 6 '17 at 10:28
Marc van LeeuwenMarc van Leeuwen
87.9k5111226
answered Jul 6 '17 at 10:28
Marc van LeeuwenMarc van Leeuwen
87.9k5111226
87.9k5111226
add a comment |
add a comment |
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$begingroup$
Linked with the topic web.mit.edu/18.06/www/Essays/newpaper_ver3.pdf
$endgroup$
– Widawensen
Jul 6 '17 at 10:29