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Here's an interesting review question I have: Find a nonzero matrix, so that when it is multiplied by another nonzero matrix, the zero matrix is the result. How would you go about solving this? Thanks!

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I don't understand: you want a matrix $A\neq 0$ such that for all $B\neq 0$, $AB=0$. But if you take $B=I$? –  Davide Giraudo Oct 21 '12 at 20:15
    
I want AB = 0 when A ≠ 0 and B ≠ 0. B can be any non-zero matrix –  Chris Camargo Oct 21 '12 at 20:19

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up vote 3 down vote accepted

First, I’d look for a simple example, so I’d start with $2\times 2$ matrices, say $A$ and $B$. If I make the bottom row of $A$ all zeroes, the bottom row of $AB$ will automatically be all zeroes. If I make the second column of $B$ all zeroes, the second column of $AB$ will automatically be all zeroes. At this point I have

$$\pmatrix{a&b\\0&0}\pmatrix{c&0\\d&0}=\pmatrix{ac+bd&0\\0&0}\;,$$

and all that’s needed in order to finish the job is to find $a,b,c$, and $d$ so that $ab+cd=0$, at least one of $a$ and $b$ is non-zero, and at least one of $c$ and $d$ is non-zero; this is very easy to do.

If you’re comfortable thinking of matrices as linear transformations, you can look for a pair of non-zero linear transformations $S$ and $T$ such that $T\circ S$ sends everything to $0$. Here again I’d try to keep it simple, so I’d look for linear transformations sending $R^2$ to itself. What if $S$ is the linear transformation that projects the plane to the $x$-axis, and $T$ is the one that projects the plane to the $y$-axis? Clearly $T\circ S$ (and for that matter $S\circ T$) sends everything to the origin, so it’s the zero transformation, but neither $S$ nor $T$ by itself does so. Having seen that, all I’d have to do is write down the matrices of $S$ and $T$.

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Thanks for your help! –  Chris Camargo Oct 21 '12 at 20:36
    
@Chris: You’re welcome! –  Brian M. Scott Oct 21 '12 at 20:36

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