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I'm practicing a problem that I came across, it's

$\text{G}$, $\text{H}$, and $\text{K}$ are $4\times4$ matrices. $\text{GHK}=0$ and $\text{H}$ is invertible, does $\text{GK}=0$?

I was trying to think in terms of ranks, if $\text{H}$ is rank $4$, then if $\text{G}$ is rank $3$, and then $\text{GH}$ is rank $3$. If $\text{K}$ happens to be rank $1$ and all in the kernel of $\text{GH}$, then $\text{GHK}=0$. But that doesn't necessarily mean that $\text{K}$ is the kernel of $\text{G}$?

In a more general case, if you a matrix without a kernel, say $\text{A}$, and you multiply it by another matrix without a kernel, you'll always get an invertible matrix? But if you multiply $\text{A}$ by a matrix with a kernel, your result is no longer valid in terms of invertibility?

Thanks!

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I'll do an example with $2\times 2$ matrices. Let $$G =K= \left(\begin{matrix}1 & 0 \\ 0 & 0 \end{matrix}\right) \quad \text{and} \quad H=\left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right).$$ We find that $GHK =0$, while $$GK= \left(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}\right).$$ What we've done is use the permutation matrix $H$ to move around the eigenvalues of $G$ and $K$. With appropriate movement, we can change the dimension of the null space.

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  • $\begingroup$ Ah, could you elaborate a little more on permutation matrices or give me a link to go to to learn more about them? They definitely haven't been covered in class but I feel as if they're useful for some of the counterexamples like this, thanks! $\endgroup$ – google1254 May 14 '13 at 15:10
  • $\begingroup$ There's not much more to be said. Permutation matrices help rearrange the blocks of a matrix. In particular, they can move around zero rows to encourage cancellation. $\endgroup$ – awwalker May 15 '13 at 16:38
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$$ \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \left( \begin{array}{rrrr} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right) \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) $$ is the zero matrix.

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  • $\begingroup$ ah, that's great! How did you come up with the matrices? $\endgroup$ – google1254 May 14 '13 at 5:06
  • $\begingroup$ @google1254 The central matrix is a permutation matrix (see my answer for more context on this). $\endgroup$ – awwalker May 14 '13 at 5:07
  • $\begingroup$ @google1254, I made it up. $\endgroup$ – Will Jagy May 14 '13 at 5:07

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