Eigenvalues and eigenvector

Question

I have found this one but do not got any clue how to solve it.

Let be $A$ a symmetrical matrix $3X3$ with real numbers. Prove that if $\lambda_1$,$\lambda_2$ are different eigenvalues with their own eigenvector $v_1\neq 0$ and $v_2\neq 0$. Then $v_1$ is orthogonal to $v_2$ from that show that $v_1$ and $v_2$ are linear independences.

To me its pretty obvious the second part but I can not prove that $v_1$,$v_2$ are orthogonal

Answer 1

Calculate the inner product

$$ \langle A v_1 | v_2 \rangle = \lambda_1 \langle v_1 | v_2\rangle \tag{1} $$

But on the other hand

$$ \langle A v_1 | v_2 \rangle = \langle v_1 | A^T v_2 \rangle = \langle v_1 | \lambda_2 v_2 \rangle = \lambda_2 \langle v_1 | v_2 \rangle \tag{2} $$

From (1) and (2) you get to the conclusion

$$ (\lambda_1 - \lambda_2) \langle v_1 | v_2 \rangle = 0 $$

Since the eigenvalues are different

$$ \langle v_1 | v_2 \rangle = 0 $$

Answer 2

I think you probably want $A$ with real coefficients here.

Suppose $Av_1=\lambda_1v_1$ and $Av_2=\lambda_2v_2$ with $\lambda_1 \neq \lambda_2$ and $A$ is symmetric $A^T=A$.

Then $$\lambda_1 v_1\cdot v_2= (Av_1)\cdot v_2 = (Av_1)^Tv_2=v_1A^Tv_2=v_1\cdot (Av_2) = \lambda_2 v_1\cdot v_2$$

Thus $\lambda_1 v_1\cdot v_2 = \lambda_2 v_1\cdot v_2$ if $v_1 \cdot v_2 \neq 0$ then you have $\lambda_1=\lambda_2$ a contradiction