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Suppose $A_1$, $A_2$ and $A_3$ are $3$ by $3$ matrices of rank $2$ such that their kernels are linearly independent. Is the following true?

Define: $V_1=A_2A_1$, $V_2=A_3A_2$ and $V_3=A_1A_3$. Then each $V_i$ is rank $1$ and $V_iV_j=0$ for $i\neq j$.

Thanks for your help!

Stan

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Did you try anything? as it is only a 3 times 3 matrix you should be able to brute force it –  Dominic Michaelis Mar 4 '13 at 13:39
    
So my thoughts so far is that for this to be true $V_1$ should have two independent vectors in it's kernel. The kernel of $A_1$ is certainly one of these and something like $A_1^{-1}ker(A_2)$ would be the other but it's not clear to me that it's possible to use the inverse of $A_1$ in the definition seeing as $A_1$ is by definition singular. Does that make sense? –  Stan Mar 4 '13 at 13:46
    
What does it mean for the kernels of matrices (linear transformationms) to be "linearly independent"?? –  DonAntonio Mar 4 '13 at 13:50
    
@DonAntonio I probably haven't been accurate in my terminology but I mean something like: Let $k_1$, $k_2$ and $k_3$ be vectors in the respective kernels, then these vectors are linearly independent. –  Stan Mar 4 '13 at 13:57
    
The kernels of $V_i$ are in general not linear independent, they can be the same as in my example if you like my answer you can upvote it :) –  Dominic Michaelis Mar 4 '13 at 14:22

1 Answer 1

up vote 2 down vote accepted

If I interpret the question correctly the assumption is wrong, take \begin{align*} A_1&= \begin{pmatrix} 0& 1 & 0\\ 0&0 &1\\ 0 & 0 & 0\\ \end{pmatrix} \qquad \operatorname{kernel} \begin{pmatrix} \alpha \\0 \\ 0 \end{pmatrix} \\ A_2&= \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 &0 \\ 0 & 0 &0 \end{pmatrix} \qquad \operatorname{kernel} \begin{pmatrix} 0 \\ 0 \\ \beta \\ \end{pmatrix}\\ A_3&= \begin{pmatrix} 0 & 0& 1\\ 0& 0 & 0 \\ 1 & 0 & 0\end{pmatrix} \qquad \operatorname{kernel}\begin{pmatrix} 0 \\\gamma \\ 0\\ \end{pmatrix} \end{align*} Than $V_1 V_2= A_2 A_1^2 A_3$ which is \begin{align*} V_1V_2&= \begin{pmatrix} 1 & 0 &0\\ 1 & 1 &0 \\ 0 & 0 & 0 \\ \end{pmatrix} \cdot \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0& 0 & 0\\ \end{pmatrix}\cdot \begin{pmatrix} 0 & 0 & 1 \\0 & 0 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 & 0 \\1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \end{align*} checked with mathematica, so we have shown that $V_i V_j=0$ for $i \neq j$ is wrong, we will show what else won't work with this example. $$V_1=A_2 A_1 = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{pmatrix} $$ So rank of $V_1$ is 2 not 1. For $V_2$ we have $$A_3 A_2 = \begin{pmatrix} 0 & 0 & 0 \\0 & 0 & 0\\ 1 & 0 & 0 \end{pmatrix} $$ Here it works, and even for $V_3$ we have $$A_1 A_3 = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\0 & 0 & 0 \\ \end{pmatrix} $$

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