I've been stuck on this for a while:

Show that a square matrixAhas no inverse when each row ofAsums to 0.

Any help would be appreciated!

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- Oct 31st 2009, 04:00 PMBrownianManMatrix Inverse question
I've been stuck on this for a while:

Show that a square matrix*A*has no inverse when each row of*A*sums to 0.

Any help would be appreciated! - Oct 31st 2009, 04:16 PMtonio

An idea: induction on n = number of rows: for n= 1, 2 is almost immediate, so assume for n: now begin simplifying your matrix by Gauss making zeros on the first column (Note, or in fact: prove, that after this is done all the rows STILL sum to zero!) except perhaps the entry 1-1 and now develop by minors wrt the first column...

Tonio - Nov 1st 2009, 12:59 AMmath2009
$\displaystyle A\begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, span(\begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix}) \in \ker(A) \neq \phi \ ,\ \therefore A $ is noninvertible