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October 19th, 2019, 06:02 PM  #1 
Newbie Joined: Oct 2019 From: Taiwan Posts: 2 Thanks: 0  Prove ABA' is singular
A is MxN matrix. B is NxN matrix. If N<M Prove ABA' is singular.(A' is transpose of A) No more conditions. Last edited by wrxue; October 19th, 2019 at 06:25 PM. 
October 19th, 2019, 11:16 PM  #2 
Senior Member Joined: Oct 2018 From: USA Posts: 102 Thanks: 77 Math Focus: Algebraic Geometry 
We know that $A$ is $m \times n$ and $B$ is $n \times n$, so $AB$ is $m \times n$ and $ABA^{t}$ is $m \times m$. Since $n<m$, $rank(A) \leq n$ and $rank(B) \leq n$ We know that $rank(AB) \leq min\{rank(A), rank(B)\}\leq n$, so $rank(AB) \leq n$ We know that $rank(A^t) = rank(A) \leq n$ Therefore, $rank(ABA^t) \leq min\{rank(AB), rank(A^t)\} \leq n$, so $rank(ABA^t) \leq n$. $ABA^t$ is $m \times m$, but $rank(ABA^t) \leq n < m$ so $ABA^t$ is singular. 
October 20th, 2019, 12:50 AM  #3 
Newbie Joined: Oct 2019 From: Taiwan Posts: 2 Thanks: 0 
Thank you!! Very clear. 

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