null space of a matrix product is a subspace of the second matrix
$begingroup$
I was given a following statement and was told it is true, but I can't figure out why. Here it is.
Let's say we have a following matrix product, $ C=AB $ Null space of $C$ is a subspace of null space of $B$.
Can someone explain why this is the case. I can't figure it out.
linear-algebra matrices vector-spaces
asked Feb 17 '17 at 15:09
flashburnflashburn
21338
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add a comment |
$begingroup$
I was given a following statement and was told it is true, but I can't figure out why. Here it is.
Let's say we have a following matrix product, $ C=AB $ Null space of $C$ is a subspace of null space of $B$.
Can someone explain why this is the case. I can't figure it out.
linear-algebra matrices vector-spaces
asked Feb 17 '17 at 15:09
flashburnflashburn
21338
$endgroup$
add a comment |
$begingroup$
I was given a following statement and was told it is true, but I can't figure out why. Here it is.
Let's say we have a following matrix product, $ C=AB $ Null space of $C$ is a subspace of null space of $B$.
Can someone explain why this is the case. I can't figure it out.
linear-algebra matrices vector-spaces
asked Feb 17 '17 at 15:09
flashburnflashburn
21338
$endgroup$
I was given a following statement and was told it is true, but I can't figure out why. Here it is.
Let's say we have a following matrix product, $ C=AB $ Null space of $C$ is a subspace of null space of $B$.
Can someone explain why this is the case. I can't figure it out.
linear-algebra matrices vector-spaces
linear-algebra matrices vector-spaces
asked Feb 17 '17 at 15:09
flashburnflashburn
21338
asked Feb 17 '17 at 15:09
flashburnflashburn
21338
asked Feb 17 '17 at 15:09
flashburnflashburn
21338
asked Feb 17 '17 at 15:09
flashburnflashburn
21338
asked Feb 17 '17 at 15:09
flashburnflashburn
21338
21338
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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As Andreas Caranti points out in his answer, the statement is not true, and in fact it’s the other way around: $operatorname{Null}(B)subseteqoperatorname{Null}(C)$. It’s not hard to see why the null space can only grow, not shrink, when you multiply by another matrix. Multiplying the zero vector by anything gives the zero vector, so if $B$ sends a vector to zero, it stays there when you multiply by $A$. So, multiplying by $A$ can add more vectors to the total null space of the product, but it can’t remove any.
edited Apr 13 '17 at 12:21
Community♦
1
answered Feb 17 '17 at 19:58
amdamd
29.5k21050
$endgroup$
$begingroup$
I'm not sure I understand. B contains vectors that make A vectors equal to 0. How is it related to matrix C, what is the connection?
$endgroup$
– flashburn
Feb 17 '17 at 21:45
1
$begingroup$
The null space is the set of vectors that are “collapsed” to zero by a matrix. Once they’ve collapsed, they stay collapsed: any matrix times zero is still zero. Multiplying the results by another matrix can only add to the set of vectors that have collapsed. It can’t get any smaller. Thus it can only be the case that $operatorname{Null}(B)subseteqoperatorname{Null}(AB)$ for any matrices $A$ and $B$.
$endgroup$
– amd
Feb 17 '17 at 23:33
add a comment |
$begingroup$
It's the other way around, as you can see by taking for instance $B$ to be the $n times n$ identity matrix (so the null space is ${ 0 }$), and $A$ to be the zero $n times n$ matrix (so that $C$ is also zero, and thus its null space is the whole space).
If $B v = 0$, then $C v = (A B) v = A (B v) = A 0 = 0$.
answered Feb 17 '17 at 15:18
Andreas CarantiAndreas Caranti
56.4k34295
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As Andreas Caranti points out in his answer, the statement is not true, and in fact it’s the other way around: $operatorname{Null}(B)subseteqoperatorname{Null}(C)$. It’s not hard to see why the null space can only grow, not shrink, when you multiply by another matrix. Multiplying the zero vector by anything gives the zero vector, so if $B$ sends a vector to zero, it stays there when you multiply by $A$. So, multiplying by $A$ can add more vectors to the total null space of the product, but it can’t remove any.
edited Apr 13 '17 at 12:21
Community♦
1
answered Feb 17 '17 at 19:58
amdamd
29.5k21050
$endgroup$
$begingroup$
I'm not sure I understand. B contains vectors that make A vectors equal to 0. How is it related to matrix C, what is the connection?
$endgroup$
– flashburn
Feb 17 '17 at 21:45
1
$begingroup$
The null space is the set of vectors that are “collapsed” to zero by a matrix. Once they’ve collapsed, they stay collapsed: any matrix times zero is still zero. Multiplying the results by another matrix can only add to the set of vectors that have collapsed. It can’t get any smaller. Thus it can only be the case that $operatorname{Null}(B)subseteqoperatorname{Null}(AB)$ for any matrices $A$ and $B$.
$endgroup$
– amd
Feb 17 '17 at 23:33
add a comment |
$begingroup$
As Andreas Caranti points out in his answer, the statement is not true, and in fact it’s the other way around: $operatorname{Null}(B)subseteqoperatorname{Null}(C)$. It’s not hard to see why the null space can only grow, not shrink, when you multiply by another matrix. Multiplying the zero vector by anything gives the zero vector, so if $B$ sends a vector to zero, it stays there when you multiply by $A$. So, multiplying by $A$ can add more vectors to the total null space of the product, but it can’t remove any.
edited Apr 13 '17 at 12:21
Community♦
1
answered Feb 17 '17 at 19:58
amdamd
29.5k21050
$endgroup$
$begingroup$
I'm not sure I understand. B contains vectors that make A vectors equal to 0. How is it related to matrix C, what is the connection?
$endgroup$
– flashburn
Feb 17 '17 at 21:45
1
$begingroup$
The null space is the set of vectors that are “collapsed” to zero by a matrix. Once they’ve collapsed, they stay collapsed: any matrix times zero is still zero. Multiplying the results by another matrix can only add to the set of vectors that have collapsed. It can’t get any smaller. Thus it can only be the case that $operatorname{Null}(B)subseteqoperatorname{Null}(AB)$ for any matrices $A$ and $B$.
$endgroup$
– amd
Feb 17 '17 at 23:33
add a comment |
$begingroup$
As Andreas Caranti points out in his answer, the statement is not true, and in fact it’s the other way around: $operatorname{Null}(B)subseteqoperatorname{Null}(C)$. It’s not hard to see why the null space can only grow, not shrink, when you multiply by another matrix. Multiplying the zero vector by anything gives the zero vector, so if $B$ sends a vector to zero, it stays there when you multiply by $A$. So, multiplying by $A$ can add more vectors to the total null space of the product, but it can’t remove any.
edited Apr 13 '17 at 12:21
Community♦
1
answered Feb 17 '17 at 19:58
amdamd
29.5k21050
$endgroup$
As Andreas Caranti points out in his answer, the statement is not true, and in fact it’s the other way around: $operatorname{Null}(B)subseteqoperatorname{Null}(C)$. It’s not hard to see why the null space can only grow, not shrink, when you multiply by another matrix. Multiplying the zero vector by anything gives the zero vector, so if $B$ sends a vector to zero, it stays there when you multiply by $A$. So, multiplying by $A$ can add more vectors to the total null space of the product, but it can’t remove any.
edited Apr 13 '17 at 12:21
Community♦
1
answered Feb 17 '17 at 19:58
amdamd
29.5k21050
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Feb 17 '17 at 19:58
amdamd
29.5k21050
answered Feb 17 '17 at 19:58
amdamd
29.5k21050
answered Feb 17 '17 at 19:58
amdamd
29.5k21050
29.5k21050
$begingroup$
I'm not sure I understand. B contains vectors that make A vectors equal to 0. How is it related to matrix C, what is the connection?
$endgroup$
– flashburn
Feb 17 '17 at 21:45
1
$begingroup$
The null space is the set of vectors that are “collapsed” to zero by a matrix. Once they’ve collapsed, they stay collapsed: any matrix times zero is still zero. Multiplying the results by another matrix can only add to the set of vectors that have collapsed. It can’t get any smaller. Thus it can only be the case that $operatorname{Null}(B)subseteqoperatorname{Null}(AB)$ for any matrices $A$ and $B$.
$endgroup$
– amd
Feb 17 '17 at 23:33
add a comment |
$begingroup$
I'm not sure I understand. B contains vectors that make A vectors equal to 0. How is it related to matrix C, what is the connection?
$endgroup$
– flashburn
Feb 17 '17 at 21:45
1
$begingroup$
The null space is the set of vectors that are “collapsed” to zero by a matrix. Once they’ve collapsed, they stay collapsed: any matrix times zero is still zero. Multiplying the results by another matrix can only add to the set of vectors that have collapsed. It can’t get any smaller. Thus it can only be the case that $operatorname{Null}(B)subseteqoperatorname{Null}(AB)$ for any matrices $A$ and $B$.
$endgroup$
– amd
Feb 17 '17 at 23:33
$begingroup$
I'm not sure I understand. B contains vectors that make A vectors equal to 0. How is it related to matrix C, what is the connection?
$endgroup$
– flashburn
Feb 17 '17 at 21:45
$begingroup$
I'm not sure I understand. B contains vectors that make A vectors equal to 0. How is it related to matrix C, what is the connection?
$endgroup$
– flashburn
Feb 17 '17 at 21:45
1
1
$begingroup$
The null space is the set of vectors that are “collapsed” to zero by a matrix. Once they’ve collapsed, they stay collapsed: any matrix times zero is still zero. Multiplying the results by another matrix can only add to the set of vectors that have collapsed. It can’t get any smaller. Thus it can only be the case that $operatorname{Null}(B)subseteqoperatorname{Null}(AB)$ for any matrices $A$ and $B$.
$endgroup$
– amd
Feb 17 '17 at 23:33
$begingroup$
The null space is the set of vectors that are “collapsed” to zero by a matrix. Once they’ve collapsed, they stay collapsed: any matrix times zero is still zero. Multiplying the results by another matrix can only add to the set of vectors that have collapsed. It can’t get any smaller. Thus it can only be the case that $operatorname{Null}(B)subseteqoperatorname{Null}(AB)$ for any matrices $A$ and $B$.
$endgroup$
– amd
Feb 17 '17 at 23:33
add a comment |
$begingroup$
It's the other way around, as you can see by taking for instance $B$ to be the $n times n$ identity matrix (so the null space is ${ 0 }$), and $A$ to be the zero $n times n$ matrix (so that $C$ is also zero, and thus its null space is the whole space).
If $B v = 0$, then $C v = (A B) v = A (B v) = A 0 = 0$.
answered Feb 17 '17 at 15:18
Andreas CarantiAndreas Caranti
56.4k34295
$endgroup$
add a comment |
$begingroup$
It's the other way around, as you can see by taking for instance $B$ to be the $n times n$ identity matrix (so the null space is ${ 0 }$), and $A$ to be the zero $n times n$ matrix (so that $C$ is also zero, and thus its null space is the whole space).
If $B v = 0$, then $C v = (A B) v = A (B v) = A 0 = 0$.
answered Feb 17 '17 at 15:18
Andreas CarantiAndreas Caranti
56.4k34295
$endgroup$
add a comment |
$begingroup$
It's the other way around, as you can see by taking for instance $B$ to be the $n times n$ identity matrix (so the null space is ${ 0 }$), and $A$ to be the zero $n times n$ matrix (so that $C$ is also zero, and thus its null space is the whole space).
If $B v = 0$, then $C v = (A B) v = A (B v) = A 0 = 0$.
answered Feb 17 '17 at 15:18
Andreas CarantiAndreas Caranti
56.4k34295
$endgroup$
It's the other way around, as you can see by taking for instance $B$ to be the $n times n$ identity matrix (so the null space is ${ 0 }$), and $A$ to be the zero $n times n$ matrix (so that $C$ is also zero, and thus its null space is the whole space).
If $B v = 0$, then $C v = (A B) v = A (B v) = A 0 = 0$.
answered Feb 17 '17 at 15:18
Andreas CarantiAndreas Caranti
56.4k34295
edited Feb 17 '17 at 15:23
answered Feb 17 '17 at 15:18
Andreas CarantiAndreas Caranti
56.4k34295
answered Feb 17 '17 at 15:18
Andreas CarantiAndreas Caranti
56.4k34295
answered Feb 17 '17 at 15:18
Andreas CarantiAndreas Caranti
56.4k34295
56.4k34295
add a comment |
add a comment |
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