Confused with Linear Algebra Proofs
I know that this is true, but I am not sure how to prove it:
For $l×m$ matrix $A$ and $m×n$ matrix $B$, it will always be the case that $operatorname{null}(B)$ is a subset of $operatorname{null}(AB)$
My other question is:
Suppose $B$ is a square matrix of size $n$. What are two statements equivalent to $operatorname{nullity}(B) > 0$ ?
Any help is appreciated, thank you!
linear-algebra
edited Nov 28 '18 at 4:16
Thomas Shelby
1,516216
asked Nov 28 '18 at 3:18
Thomas
62
add a comment |
I know that this is true, but I am not sure how to prove it:
For $l×m$ matrix $A$ and $m×n$ matrix $B$, it will always be the case that $operatorname{null}(B)$ is a subset of $operatorname{null}(AB)$
My other question is:
Suppose $B$ is a square matrix of size $n$. What are two statements equivalent to $operatorname{nullity}(B) > 0$ ?
Any help is appreciated, thank you!
linear-algebra
edited Nov 28 '18 at 4:16
Thomas Shelby
1,516216
asked Nov 28 '18 at 3:18
Thomas
62
Rank-nullity theorem does it for you.
– Sean Roberson
Nov 28 '18 at 3:21
The first statement you prove just by checking it..
– Randall
Nov 28 '18 at 3:22
add a comment |
I know that this is true, but I am not sure how to prove it:
For $l×m$ matrix $A$ and $m×n$ matrix $B$, it will always be the case that $operatorname{null}(B)$ is a subset of $operatorname{null}(AB)$
My other question is:
Suppose $B$ is a square matrix of size $n$. What are two statements equivalent to $operatorname{nullity}(B) > 0$ ?
Any help is appreciated, thank you!
linear-algebra
edited Nov 28 '18 at 4:16
Thomas Shelby
1,516216
asked Nov 28 '18 at 3:18
Thomas
62
I know that this is true, but I am not sure how to prove it:
For $l×m$ matrix $A$ and $m×n$ matrix $B$, it will always be the case that $operatorname{null}(B)$ is a subset of $operatorname{null}(AB)$
My other question is:
Suppose $B$ is a square matrix of size $n$. What are two statements equivalent to $operatorname{nullity}(B) > 0$ ?
Any help is appreciated, thank you!
linear-algebra
linear-algebra
edited Nov 28 '18 at 4:16
Thomas Shelby
1,516216
asked Nov 28 '18 at 3:18
Thomas
62
edited Nov 28 '18 at 4:16
Thomas Shelby
1,516216
asked Nov 28 '18 at 3:18
Thomas
62
edited Nov 28 '18 at 4:16
Thomas Shelby
1,516216
edited Nov 28 '18 at 4:16
Thomas Shelby
1,516216
edited Nov 28 '18 at 4:16
Thomas Shelby
1,516216
1,516216
asked Nov 28 '18 at 3:18
Thomas
62
asked Nov 28 '18 at 3:18
Thomas
62
asked Nov 28 '18 at 3:18
Thomas
62
62
Rank-nullity theorem does it for you.
– Sean Roberson
Nov 28 '18 at 3:21
The first statement you prove just by checking it..
– Randall
Nov 28 '18 at 3:22
add a comment |
Rank-nullity theorem does it for you.
– Sean Roberson
Nov 28 '18 at 3:21
The first statement you prove just by checking it..
– Randall
Nov 28 '18 at 3:22
Rank-nullity theorem does it for you.
– Sean Roberson
Nov 28 '18 at 3:21
Rank-nullity theorem does it for you.
– Sean Roberson
Nov 28 '18 at 3:21
The first statement you prove just by checking it..
– Randall
Nov 28 '18 at 3:22
The first statement you prove just by checking it..
– Randall
Nov 28 '18 at 3:22
add a comment |
1 Answer
1
active
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First, the dimensions of the matrices are irrelevant except for the fact that we want $AB$ to be a well defined matrix product. So you can safely ignore that for the moment.
You want to prove that $null(B)subset null(AB)$. So pick a vector in $null(B)$, say $vec{v}$. What can you say about the matrix-vector product $(AB)vec{v}$?
You also want to say something about properties of a matrix with positive nullity. As the comment on your post suggests, the Rank-Nullity theorem does the trick for this task.
answered Nov 28 '18 at 3:23
Valborg
542
Could you expand on these and give examples? Sorry, I am pretty confused about how to get the answers for these...
– Thomas
Nov 28 '18 at 5:33
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
First, the dimensions of the matrices are irrelevant except for the fact that we want $AB$ to be a well defined matrix product. So you can safely ignore that for the moment.
You want to prove that $null(B)subset null(AB)$. So pick a vector in $null(B)$, say $vec{v}$. What can you say about the matrix-vector product $(AB)vec{v}$?
You also want to say something about properties of a matrix with positive nullity. As the comment on your post suggests, the Rank-Nullity theorem does the trick for this task.
answered Nov 28 '18 at 3:23
Valborg
542
Could you expand on these and give examples? Sorry, I am pretty confused about how to get the answers for these...
– Thomas
Nov 28 '18 at 5:33
add a comment |
First, the dimensions of the matrices are irrelevant except for the fact that we want $AB$ to be a well defined matrix product. So you can safely ignore that for the moment.
You want to prove that $null(B)subset null(AB)$. So pick a vector in $null(B)$, say $vec{v}$. What can you say about the matrix-vector product $(AB)vec{v}$?
You also want to say something about properties of a matrix with positive nullity. As the comment on your post suggests, the Rank-Nullity theorem does the trick for this task.
answered Nov 28 '18 at 3:23
Valborg
542
Could you expand on these and give examples? Sorry, I am pretty confused about how to get the answers for these...
– Thomas
Nov 28 '18 at 5:33
add a comment |
First, the dimensions of the matrices are irrelevant except for the fact that we want $AB$ to be a well defined matrix product. So you can safely ignore that for the moment.
You want to prove that $null(B)subset null(AB)$. So pick a vector in $null(B)$, say $vec{v}$. What can you say about the matrix-vector product $(AB)vec{v}$?
You also want to say something about properties of a matrix with positive nullity. As the comment on your post suggests, the Rank-Nullity theorem does the trick for this task.
answered Nov 28 '18 at 3:23
Valborg
542
First, the dimensions of the matrices are irrelevant except for the fact that we want $AB$ to be a well defined matrix product. So you can safely ignore that for the moment.
You want to prove that $null(B)subset null(AB)$. So pick a vector in $null(B)$, say $vec{v}$. What can you say about the matrix-vector product $(AB)vec{v}$?
You also want to say something about properties of a matrix with positive nullity. As the comment on your post suggests, the Rank-Nullity theorem does the trick for this task.
answered Nov 28 '18 at 3:23
Valborg
542
answered Nov 28 '18 at 3:23
Valborg
542
answered Nov 28 '18 at 3:23
Valborg
542
answered Nov 28 '18 at 3:23
Valborg
542
542
Could you expand on these and give examples? Sorry, I am pretty confused about how to get the answers for these...
– Thomas
Nov 28 '18 at 5:33
add a comment |
Could you expand on these and give examples? Sorry, I am pretty confused about how to get the answers for these...
– Thomas
Nov 28 '18 at 5:33
Could you expand on these and give examples? Sorry, I am pretty confused about how to get the answers for these...
– Thomas
Nov 28 '18 at 5:33
Could you expand on these and give examples? Sorry, I am pretty confused about how to get the answers for these...
– Thomas
Nov 28 '18 at 5:33
add a comment |
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Rank-nullity theorem does it for you.
– Sean Roberson
Nov 28 '18 at 3:21
The first statement you prove just by checking it..
– Randall
Nov 28 '18 at 3:22