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October 19th, 2019, 06:02 PM   #1
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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.
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October 19th, 2019, 11:16 PM   #2
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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.
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October 20th, 2019, 12:50 AM   #3
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Thank you!!
Very clear.
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