I have to prove that if matrix has two identical rows or columns then it is not a reversible matrix. I know that in such scenario matrix determinant is equal zero, but I cannot use determinants in my proof.
Can anybody give a hint?
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2@Timbuc Rather, non-invertibility. – apnorton Oct 19 '14 at 12:39
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1Does not have an inverse – Mark Bennet Oct 19 '14 at 12:41
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Depending on whether it is a row or a column, multiply by a suspected inverse on right or left.
What happens when you multiply by equal rows (or columns with reversed case) in the matrix - consider the equivalent rows in the product? Can you derive a contradiction?
Mark Bennet
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Let's do it for equal columns. Suppose the matrix $A$ has equal rows, which are the columns number $k$ and $l$. Consider the column vector $e_k-e_l$ (which has $1$ on its $k$-th row and $-1$ on its $l$-th row). Then $$ A(e_k-e_l)=Ae_k-Ae_l=0 $$ because the product of $A$ by the column $e_i$ is the $i$-th column of $A$.
Can $A$ be invertible if $Ax=0$ for some vector $x\ne0$?
For equal rows, just use the transpose.
egreg
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We know (or you can prove it) that a matrix is invertible iff when we reduce it by rows (or columns), we get a reduced form none of which rows (columns) is all zeros
In your case the above is not going to happen (why?)
Timbuc
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