Dot product of eigen vectors of square matrix

Given a $3\times 3$ matrix $A$ and column vectors $v_1,v_2$ such that

$$A^{t} = A \$$ $$Av_1 = \lambda_1v_1$$ $$Av_2 = \lambda_2v_2$$

I need to prove that $$v_1\cdot v_2=0$$ I've tried doing this $$Av_1=\lambda_1v_1\Rightarrow v_1^tA^t=\lambda_1v_1^t\Rightarrow v_1^tA=\lambda_1v_1^t$$ So that $$v_1^tAv_2=\lambda_1v_1\cdot v_2$$ But I don't think that get's me anywhere

(This is not a homework question, it's from a past paper to which I have no solutions)

• What does $A^t = A$ implies? What do you know about $v_1$ and $v_2$ if $A$ is symmetric? – j-a Jan 9 '12 at 20:36
• You have deduced the equation $v_1^tAv_2 = \lambda_1v_1\cdot v_2$. What happens if you transpose both sides of this equation? (Remember, the transpose of a number, i.e. a 1x1 matrix, is itself.) And what happens if you made the same deduction starting from the equation $Av_2 = \lambda_2v_2$? – bradhd Jan 9 '12 at 20:40

You are needing to assume that the eigenvalues are different, no? Otherwise, $A=0$ satisfies $A=A^\top$ and $Ax=0x$ as well as $A2x=0(2x)$ yet $\langle x,2x\rangle\ne0$. So, basically the idea then comes from the fact that $A^\top$ is the adjoint for $A$ on $\mathbb{R}^n$ with the usual inner product. In other words, $A^\top$ satisfies the relation $\langle Ax,y\rangle=\langle x,A^\top y\rangle$. So, since $\lambda_1\ne\lambda_2$ we may assume WLOG that $\lambda_1\ne0$. In this case we have that
\begin{aligned}\langle v_1,v_2\rangle &=\frac{1}{\lambda_1}\langle \lambda_1 v_1,v_2\rangle\\ &=\frac{1}{\lambda_1}\langle Av_1,v_2\rangle\\ &=\frac{1}{\lambda_1}\langle v_1,A^\top v_2\rangle\\ &=\frac{1}{\lambda_1}\langle v_1,Av_2\rangle\\ &=\frac{1}{\lambda_1}\langle v_1,\lambda_2 v_2\rangle\\\ &=\frac{\lambda_2}{\lambda_1}\langle v_1,v_2\rangle\end{aligned}
and since $\lambda_2\ne\lambda_1$ this implies that $\langle v_1,v_2\rangle=0$.