Proving Singular Matrix A Has Nonzero Matrix B: Linear Algebra Problem

Braka
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The problem is prove that for every square singular matrix A there is a nonzero matrix B, such that AB equals the zero matrix.

I got AB to equal the idenity matrix, but have no clue how to get it to the zero matrix.
 
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If A is viewed as the matrix of a linear transformation, then it being singular is the same as saying that it maps a subspace in the domain to the 0 subspace of the range. Find the kernel of A and let B be a matrix of vectors in that space.
 
you said:
I got AB to equal the idenity matrix, but have no clue how to get it to the zero matrix.

how did you do that? If A is singular then it doesn't have an inverse, but you found one namely B.
 
If A is singular, det(A)= 0. Also det(AB)= det(A)det(B). You, apparently, have proved that det(AB)= 0(det(B)= 0= det(I)= 1. A miracle indeed!
 

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