Confused with Linear Algebra Proofs












1














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!










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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
















1














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!










share|cite|improve this question
























  • 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














1












1








1







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!










share|cite|improve this question















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






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share|cite|improve this question













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share|cite|improve this question








edited Nov 28 '18 at 4:16









Thomas Shelby

1,830217




1,830217










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


















  • 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










1 Answer
1






active

oldest

votes


















1














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.






share|cite|improve this answer





















  • 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











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














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.






share|cite|improve this answer





















  • 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
















1














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.






share|cite|improve this answer





















  • 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














1












1








1






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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


















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