Relationship between linear transformations
$begingroup$
Suppose that for any $Ntimes 1$ vector $t$ that satisfies $At=0$ for some $2times N$ non-zero matrix $A$, that same vector necessarily satisfies $Bt=0$ for some $1times N$ non-zero matrix $B$. What is the relationship between the matrix $A$ and the matrix $B$? (i.e., if the null space of $A$ is a subset of the null space of $B$, how must $A$ be related to $B$?)
linear-algebra matrices
asked Dec 13 '18 at 20:13
MikeMike
368110
$endgroup$
add a comment |
$begingroup$
Suppose that for any $Ntimes 1$ vector $t$ that satisfies $At=0$ for some $2times N$ non-zero matrix $A$, that same vector necessarily satisfies $Bt=0$ for some $1times N$ non-zero matrix $B$. What is the relationship between the matrix $A$ and the matrix $B$? (i.e., if the null space of $A$ is a subset of the null space of $B$, how must $A$ be related to $B$?)
linear-algebra matrices
asked Dec 13 '18 at 20:13
MikeMike
368110
$endgroup$
$begingroup$
No relation in general. Just take $B=0$ and any other $A$.
$endgroup$
– Dietrich Burde
Dec 13 '18 at 20:16
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Thanks. I'll edit the question to include the conditions that $A$ and $B$ are non-zero.
$endgroup$
– Mike
Dec 13 '18 at 20:17
$begingroup$
Another way of asking the question is: if every vector $t$ that is orthogonal to vector $x$ and vector $y$ is orthogonal to vector $z$, are $x$, $y$, and $z$ related to each other in some fundamental way?
$endgroup$
– Mike
Dec 13 '18 at 20:49
add a comment |
$begingroup$
Suppose that for any $Ntimes 1$ vector $t$ that satisfies $At=0$ for some $2times N$ non-zero matrix $A$, that same vector necessarily satisfies $Bt=0$ for some $1times N$ non-zero matrix $B$. What is the relationship between the matrix $A$ and the matrix $B$? (i.e., if the null space of $A$ is a subset of the null space of $B$, how must $A$ be related to $B$?)
linear-algebra matrices
asked Dec 13 '18 at 20:13
MikeMike
368110
$endgroup$
Suppose that for any $Ntimes 1$ vector $t$ that satisfies $At=0$ for some $2times N$ non-zero matrix $A$, that same vector necessarily satisfies $Bt=0$ for some $1times N$ non-zero matrix $B$. What is the relationship between the matrix $A$ and the matrix $B$? (i.e., if the null space of $A$ is a subset of the null space of $B$, how must $A$ be related to $B$?)
linear-algebra matrices
linear-algebra matrices
asked Dec 13 '18 at 20:13
MikeMike
368110
asked Dec 13 '18 at 20:13
MikeMike
368110
edited Dec 13 '18 at 20:17
Mike
asked Dec 13 '18 at 20:13
MikeMike
368110
asked Dec 13 '18 at 20:13
MikeMike
368110
asked Dec 13 '18 at 20:13
MikeMike
368110
368110
$begingroup$
No relation in general. Just take $B=0$ and any other $A$.
$endgroup$
– Dietrich Burde
Dec 13 '18 at 20:16
$begingroup$
Thanks. I'll edit the question to include the conditions that $A$ and $B$ are non-zero.
$endgroup$
– Mike
Dec 13 '18 at 20:17
$begingroup$
Another way of asking the question is: if every vector $t$ that is orthogonal to vector $x$ and vector $y$ is orthogonal to vector $z$, are $x$, $y$, and $z$ related to each other in some fundamental way?
$endgroup$
– Mike
Dec 13 '18 at 20:49
add a comment |
$begingroup$
No relation in general. Just take $B=0$ and any other $A$.
$endgroup$
– Dietrich Burde
Dec 13 '18 at 20:16
$begingroup$
Thanks. I'll edit the question to include the conditions that $A$ and $B$ are non-zero.
$endgroup$
– Mike
Dec 13 '18 at 20:17
$begingroup$
Another way of asking the question is: if every vector $t$ that is orthogonal to vector $x$ and vector $y$ is orthogonal to vector $z$, are $x$, $y$, and $z$ related to each other in some fundamental way?
$endgroup$
– Mike
Dec 13 '18 at 20:49
$begingroup$
No relation in general. Just take $B=0$ and any other $A$.
$endgroup$
– Dietrich Burde
Dec 13 '18 at 20:16
$begingroup$
No relation in general. Just take $B=0$ and any other $A$.
$endgroup$
– Dietrich Burde
Dec 13 '18 at 20:16
$begingroup$
Thanks. I'll edit the question to include the conditions that $A$ and $B$ are non-zero.
$endgroup$
– Mike
Dec 13 '18 at 20:17
$begingroup$
Thanks. I'll edit the question to include the conditions that $A$ and $B$ are non-zero.
$endgroup$
– Mike
Dec 13 '18 at 20:17
$begingroup$
Another way of asking the question is: if every vector $t$ that is orthogonal to vector $x$ and vector $y$ is orthogonal to vector $z$, are $x$, $y$, and $z$ related to each other in some fundamental way?
$endgroup$
– Mike
Dec 13 '18 at 20:49
$begingroup$
Another way of asking the question is: if every vector $t$ that is orthogonal to vector $x$ and vector $y$ is orthogonal to vector $z$, are $x$, $y$, and $z$ related to each other in some fundamental way?
$endgroup$
– Mike
Dec 13 '18 at 20:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$Bt=0implies t$ is orthogonal to $B$. As for $operatorname{ker}A$, we have $operatorname{ker}Asubset operatorname{ker}B=B^{perp}$.
answered Dec 13 '18 at 20:49
Chris CusterChris Custer
13.9k3827
$endgroup$
$begingroup$
Thanks. If $ker A subset ker B$, what does this tell us about the relationship between the row vectors in A and the vector B? Is there some linear dependence among them?
$endgroup$
– Mike
Dec 13 '18 at 21:16
$begingroup$
@Mike The only thing that you can say for sure is that they’re all elements of the orthogonal complement of the span of all of the $t$’s
$endgroup$
– amd
Dec 13 '18 at 21:37
$begingroup$
The row space of $A$ contains that of $B$. So there's dependence.
$endgroup$
– Chris Custer
Dec 13 '18 at 22:06
add a comment |
$begingroup$
If $t$ and $B$ are non-zero, write $B cdot t$ as $langle v, t rangle$, then it is clear that any vector $v$ in the $N-1$ dimensional space perpendicular to t, i.e. $t^perp$, will satisfy $langle v, t rangle = 0$.
Suppose $N gt 3$. Since you can choose the lines of A to be any vector in $t^perp$, and $t^perp$ is $N-1$ dimensional, there is no necessary relation between the null spaces of $A$ and $B$, i.e., the null space of $A$ may or may not contain the null space of $B$.
if $N = 3$, then either the two lines of $A$ are linearly independent, or they are not. If they are linearly independent, they span $t^perp$, so the null space of $A$ contains the null space of $B$. If they are not linearly independent, they either the null space of $A$ is identical to the null space of $B$, or they are different.
Finally, if $N = 2$, then $t^perp$ is one dimensional, and both lines of $A$ must be proportional to $v$, so the null space of both $A$ and $B$ must be the same.
answered Dec 14 '18 at 12:31
Marcelo Roberto JimenezMarcelo Roberto Jimenez
464
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
$Bt=0implies t$ is orthogonal to $B$. As for $operatorname{ker}A$, we have $operatorname{ker}Asubset operatorname{ker}B=B^{perp}$.
answered Dec 13 '18 at 20:49
Chris CusterChris Custer
13.9k3827
$endgroup$
$begingroup$
Thanks. If $ker A subset ker B$, what does this tell us about the relationship between the row vectors in A and the vector B? Is there some linear dependence among them?
$endgroup$
– Mike
Dec 13 '18 at 21:16
$begingroup$
@Mike The only thing that you can say for sure is that they’re all elements of the orthogonal complement of the span of all of the $t$’s
$endgroup$
– amd
Dec 13 '18 at 21:37
$begingroup$
The row space of $A$ contains that of $B$. So there's dependence.
$endgroup$
– Chris Custer
Dec 13 '18 at 22:06
add a comment |
$begingroup$
$Bt=0implies t$ is orthogonal to $B$. As for $operatorname{ker}A$, we have $operatorname{ker}Asubset operatorname{ker}B=B^{perp}$.
answered Dec 13 '18 at 20:49
Chris CusterChris Custer
13.9k3827
$endgroup$
$begingroup$
Thanks. If $ker A subset ker B$, what does this tell us about the relationship between the row vectors in A and the vector B? Is there some linear dependence among them?
$endgroup$
– Mike
Dec 13 '18 at 21:16
$begingroup$
@Mike The only thing that you can say for sure is that they’re all elements of the orthogonal complement of the span of all of the $t$’s
$endgroup$
– amd
Dec 13 '18 at 21:37
$begingroup$
The row space of $A$ contains that of $B$. So there's dependence.
$endgroup$
– Chris Custer
Dec 13 '18 at 22:06
add a comment |
$begingroup$
$Bt=0implies t$ is orthogonal to $B$. As for $operatorname{ker}A$, we have $operatorname{ker}Asubset operatorname{ker}B=B^{perp}$.
answered Dec 13 '18 at 20:49
Chris CusterChris Custer
13.9k3827
$endgroup$
$Bt=0implies t$ is orthogonal to $B$. As for $operatorname{ker}A$, we have $operatorname{ker}Asubset operatorname{ker}B=B^{perp}$.
answered Dec 13 '18 at 20:49
Chris CusterChris Custer
13.9k3827
answered Dec 13 '18 at 20:49
Chris CusterChris Custer
13.9k3827
answered Dec 13 '18 at 20:49
Chris CusterChris Custer
13.9k3827
answered Dec 13 '18 at 20:49
Chris CusterChris Custer
13.9k3827
13.9k3827
$begingroup$
Thanks. If $ker A subset ker B$, what does this tell us about the relationship between the row vectors in A and the vector B? Is there some linear dependence among them?
$endgroup$
– Mike
Dec 13 '18 at 21:16
$begingroup$
@Mike The only thing that you can say for sure is that they’re all elements of the orthogonal complement of the span of all of the $t$’s
$endgroup$
– amd
Dec 13 '18 at 21:37
$begingroup$
The row space of $A$ contains that of $B$. So there's dependence.
$endgroup$
– Chris Custer
Dec 13 '18 at 22:06
add a comment |
$begingroup$
Thanks. If $ker A subset ker B$, what does this tell us about the relationship between the row vectors in A and the vector B? Is there some linear dependence among them?
$endgroup$
– Mike
Dec 13 '18 at 21:16
$begingroup$
@Mike The only thing that you can say for sure is that they’re all elements of the orthogonal complement of the span of all of the $t$’s
$endgroup$
– amd
Dec 13 '18 at 21:37
$begingroup$
The row space of $A$ contains that of $B$. So there's dependence.
$endgroup$
– Chris Custer
Dec 13 '18 at 22:06
$begingroup$
Thanks. If $ker A subset ker B$, what does this tell us about the relationship between the row vectors in A and the vector B? Is there some linear dependence among them?
$endgroup$
– Mike
Dec 13 '18 at 21:16
$begingroup$
Thanks. If $ker A subset ker B$, what does this tell us about the relationship between the row vectors in A and the vector B? Is there some linear dependence among them?
$endgroup$
– Mike
Dec 13 '18 at 21:16
$begingroup$
@Mike The only thing that you can say for sure is that they’re all elements of the orthogonal complement of the span of all of the $t$’s
$endgroup$
– amd
Dec 13 '18 at 21:37
$begingroup$
@Mike The only thing that you can say for sure is that they’re all elements of the orthogonal complement of the span of all of the $t$’s
$endgroup$
– amd
Dec 13 '18 at 21:37
$begingroup$
The row space of $A$ contains that of $B$. So there's dependence.
$endgroup$
– Chris Custer
Dec 13 '18 at 22:06
$begingroup$
The row space of $A$ contains that of $B$. So there's dependence.
$endgroup$
– Chris Custer
Dec 13 '18 at 22:06
add a comment |
$begingroup$
If $t$ and $B$ are non-zero, write $B cdot t$ as $langle v, t rangle$, then it is clear that any vector $v$ in the $N-1$ dimensional space perpendicular to t, i.e. $t^perp$, will satisfy $langle v, t rangle = 0$.
Suppose $N gt 3$. Since you can choose the lines of A to be any vector in $t^perp$, and $t^perp$ is $N-1$ dimensional, there is no necessary relation between the null spaces of $A$ and $B$, i.e., the null space of $A$ may or may not contain the null space of $B$.
if $N = 3$, then either the two lines of $A$ are linearly independent, or they are not. If they are linearly independent, they span $t^perp$, so the null space of $A$ contains the null space of $B$. If they are not linearly independent, they either the null space of $A$ is identical to the null space of $B$, or they are different.
Finally, if $N = 2$, then $t^perp$ is one dimensional, and both lines of $A$ must be proportional to $v$, so the null space of both $A$ and $B$ must be the same.
answered Dec 14 '18 at 12:31
Marcelo Roberto JimenezMarcelo Roberto Jimenez
464
$endgroup$
add a comment |
$begingroup$
If $t$ and $B$ are non-zero, write $B cdot t$ as $langle v, t rangle$, then it is clear that any vector $v$ in the $N-1$ dimensional space perpendicular to t, i.e. $t^perp$, will satisfy $langle v, t rangle = 0$.
Suppose $N gt 3$. Since you can choose the lines of A to be any vector in $t^perp$, and $t^perp$ is $N-1$ dimensional, there is no necessary relation between the null spaces of $A$ and $B$, i.e., the null space of $A$ may or may not contain the null space of $B$.
if $N = 3$, then either the two lines of $A$ are linearly independent, or they are not. If they are linearly independent, they span $t^perp$, so the null space of $A$ contains the null space of $B$. If they are not linearly independent, they either the null space of $A$ is identical to the null space of $B$, or they are different.
Finally, if $N = 2$, then $t^perp$ is one dimensional, and both lines of $A$ must be proportional to $v$, so the null space of both $A$ and $B$ must be the same.
answered Dec 14 '18 at 12:31
Marcelo Roberto JimenezMarcelo Roberto Jimenez
464
$endgroup$
add a comment |
$begingroup$
If $t$ and $B$ are non-zero, write $B cdot t$ as $langle v, t rangle$, then it is clear that any vector $v$ in the $N-1$ dimensional space perpendicular to t, i.e. $t^perp$, will satisfy $langle v, t rangle = 0$.
Suppose $N gt 3$. Since you can choose the lines of A to be any vector in $t^perp$, and $t^perp$ is $N-1$ dimensional, there is no necessary relation between the null spaces of $A$ and $B$, i.e., the null space of $A$ may or may not contain the null space of $B$.
if $N = 3$, then either the two lines of $A$ are linearly independent, or they are not. If they are linearly independent, they span $t^perp$, so the null space of $A$ contains the null space of $B$. If they are not linearly independent, they either the null space of $A$ is identical to the null space of $B$, or they are different.
Finally, if $N = 2$, then $t^perp$ is one dimensional, and both lines of $A$ must be proportional to $v$, so the null space of both $A$ and $B$ must be the same.
answered Dec 14 '18 at 12:31
Marcelo Roberto JimenezMarcelo Roberto Jimenez
464
$endgroup$
If $t$ and $B$ are non-zero, write $B cdot t$ as $langle v, t rangle$, then it is clear that any vector $v$ in the $N-1$ dimensional space perpendicular to t, i.e. $t^perp$, will satisfy $langle v, t rangle = 0$.
Suppose $N gt 3$. Since you can choose the lines of A to be any vector in $t^perp$, and $t^perp$ is $N-1$ dimensional, there is no necessary relation between the null spaces of $A$ and $B$, i.e., the null space of $A$ may or may not contain the null space of $B$.
if $N = 3$, then either the two lines of $A$ are linearly independent, or they are not. If they are linearly independent, they span $t^perp$, so the null space of $A$ contains the null space of $B$. If they are not linearly independent, they either the null space of $A$ is identical to the null space of $B$, or they are different.
Finally, if $N = 2$, then $t^perp$ is one dimensional, and both lines of $A$ must be proportional to $v$, so the null space of both $A$ and $B$ must be the same.
answered Dec 14 '18 at 12:31
Marcelo Roberto JimenezMarcelo Roberto Jimenez
464
answered Dec 14 '18 at 12:31
Marcelo Roberto JimenezMarcelo Roberto Jimenez
464
answered Dec 14 '18 at 12:31
Marcelo Roberto JimenezMarcelo Roberto Jimenez
464
answered Dec 14 '18 at 12:31
Marcelo Roberto JimenezMarcelo Roberto Jimenez
464
464
add a comment |
add a comment |
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$begingroup$
No relation in general. Just take $B=0$ and any other $A$.
$endgroup$
– Dietrich Burde
Dec 13 '18 at 20:16
$begingroup$
Thanks. I'll edit the question to include the conditions that $A$ and $B$ are non-zero.
$endgroup$
– Mike
Dec 13 '18 at 20:17
$begingroup$
Another way of asking the question is: if every vector $t$ that is orthogonal to vector $x$ and vector $y$ is orthogonal to vector $z$, are $x$, $y$, and $z$ related to each other in some fundamental way?
$endgroup$
– Mike
Dec 13 '18 at 20:49