Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

up vote 1 down vote accepted

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.

share|improve this answer
1  
$B$ 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
add comment

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.

share|improve this answer
    
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
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.