# A B matrices and Av Bv dependent vectors

A,B nxn complex matrices : Prove that exist a vector v(not 0), that A(v) and B(v) are dependent.

Extra question:

What if A,B are real matrices?

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If $\det B=0$, then the proof is trivial. If $\det B\ne 0$, then it boils down to prove that $\exists z\in\mathbb C,\exists v\in \mathbb C^n:$ $(A-zB)v=0$, which is equivalent to prove that $\exists z\in\mathbb C$: $\det (A-zB)=0$ or even further, $\exists z\in\mathbb C$: $\det (AB^{-1}-zI)=0$. Clearly, this polynomial has roots in $\mathbb C$, so we can conclude the proof.

If we want to work only in $\mathbb R$, then we can build a counterexample for even dimensions:

$$A=\begin{pmatrix}1&0\\0&1\end{pmatrix},\quad B=\begin{pmatrix}0&1\\-1&0\end{pmatrix}.$$ $Av=v$, but $B$ doesn't have any eigenvectors in $\mathbb R^2$.

Edit

In the case of odd dimensions, however, the hypothesys holds. Indeed, let's take our reasoning for complex case, replace $\mathbb C$ by $\mathbb R$ everywhere until the part $\exists z\in\mathbb R$: $\det (AB^{-1}-zI)=0$. In the odd-dimensioned space this determinant is a polynomial of the odd order, hence it has roots in $\mathbb R$ and thus we can find such $v$ that $Av$ and $Bv$ are dependant.

To summarize:

1. Complex case. Such $v$ exists.

2. Real case, odd dimension. Such $v$ exists.

3. Real case, even dimension. Depends on matrices, we can give examples when such $v$ exists and when it does not.

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$B$ may be zero, and $A$ nonsingular, in which case the $\det$ has no root in $\mathbb{C}$ :-). – copper.hat Jul 3 '13 at 17:07
@copper.hat I'll edit my post to take this into account. – TZakrevskiy Jul 3 '13 at 17:09
I think that for n=2k+1 it works in the real field. But i don't know show to prove it. – mrprottolo Jul 3 '13 at 19:15
@mrprottolo see edit. – TZakrevskiy Jul 3 '13 at 20:05
+1: Nice, complete answer. – copper.hat Jul 4 '13 at 13:01

If $A$ is singular, then $Av=0$ for some $v$, and hence $Bv, Av$ are dependent.

Similarly if $B$ is singular.

If $A,B$ are nonsingular, then $A^{-1}B$ has some non-zero eigenvalue $\lambda$ and eigenvector $v$. Then $A^{-1}Bv = \lambda v$, or $Bv = \lambda A v$, hence $Bv, Av$ are dependent. (Thanks to TZakrevskiy for catching an earlier error here.)

Note: By considering $A=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ and $B=\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$, and looking for a solution to $(A-\lambda B) v = 0$, we see that we must have $\lambda = 0$, so, in general, we cannot find a $v$ such that both $Av,Bv$ are non-zero as well.

Addendum: Answer to real field case: Take $A=I$ and $B=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$, and suppose $Av, Bv$ are dependent. Since $v \neq 0$ and $B$ is nonsingular, we have $Bv \neq 0$, so we can write $Av = \lambda Bv$ for some $\lambda \in \mathbb{R}$. However, this imples $Av = v = \lambda Bv$, which would mean that $\frac{1}{\lambda}$ is an eigenvalue of $B$, which is contradiction, since $B$ has no real eigenvalues. So no such $v$ exists in this case.

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Did you mean $Bv=\lambda Av$? – TZakrevskiy Jul 3 '13 at 17:07
@TZakrevskiy: Good catch :-). – copper.hat Jul 3 '13 at 17:07
@TZakrevskiy: Why delete your answer, it's a perfectly good way of dealing with $A$ non singular, and the singular case can be dealt with separately. Alternative approaches are always good. – copper.hat Jul 3 '13 at 17:15
That was tricky. Thanks so much, I was going crazy with that. :) – mrprottolo Jul 3 '13 at 17:15
Just added the second part of the problem. :) – mrprottolo Jul 3 '13 at 17:30