Let A be an m x n matrix, where m<n, then A^t(A) is an n by n matrix. show that det(A^t(A))=0

also if A is a 2 x 3 matrix can you say that det (AA^t)=0

thanks

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- Apr 24th 2009, 07:01 PMb0mb3rzmatrix and determinants
Let A be an m x n matrix, where m<n, then A^t(A) is an n by n matrix. show that det(A^t(A))=0

also if A is a 2 x 3 matrix can you say that det (AA^t)=0

thanks - Apr 24th 2009, 07:13 PMJameson
I'm not the best at Linear Algebra but I'll give this a crack until someone else helps you out.

I'm assuming A^t(A) to mean the transpose matrix of A. So if A is an m x n matrix then the transpose should be n x m, since the rows and columns switch.

Before I try to help, is this what you meant? - Apr 24th 2009, 07:18 PMb0mb3rz
yeah it is and it says hint, consider the system of homogeneous linear equations Ax=0

- Apr 24th 2009, 07:33 PMJameson
- Apr 24th 2009, 08:40 PMNonCommAlg
**b0mb3rz**didn't probably make it clear but i think what he meant was $\displaystyle \det (A^TA)=0.$ this is not difficult to prove: $\displaystyle \text{rank}(A^TA) \leq \min \ (\text{rank}A^T, \text{rank}A) \leq \min \ (m,n)=m < n. \ \ \ \color{red}(*)$

thus, by the rank-nullity theorem, the null-space of $\displaystyle A^TA$ is non-zero and hence $\displaystyle \det (A^TA) = 0.$

$\displaystyle \color{red}(*)$ you may also argue that: $\displaystyle \text{rank}(A^TA)=\text{rank}A \leq \min(m,n)=m < n.$