# Dot product of the eigenvectors of symmetric positive definite matrix is?

The value of the dot product of the eigenvectors corresponding to any pair of different eigenvalues of a $4−by−4$ symmetric positive definite matrix is _____ .

### My attempt:

Suppose $λ_1$ and $λ_2$ are distinct eigenvectors of $A$, with corresponding eigenvectors , $x$ and $y$ respectively. consider $(Qx) · (Qy)$ for two eigenvectors $x$ and $y$ with different eigenvalues .

Then we have

$(Qx) · (Qy) = (λ_1x) · (λ_2y) = λ¯_1λ_2x · y$

and

$(Qx) · (Qy) = x^HQ^HQy = x^Hy = x · y.$

Compare the above two equations, we have

$λ¯_1λ_2x · y = x · y.$

since $|λ_1|^2 = λ¯_1λ_1 = 1$, thus $λ¯_1 = 1/λ_1$.

This implies $λ¯_1λ_2 = λ_2/λ_1 ≠ 1$,

since $λ_1$ are $λ_2$ are distinct. It follows that $x · y = 0$, i.e. the eigenvectors $x$ and $y$ are orthogonal.

Whether my solution is correct or not ?

Let $\lambda_1\neq\lambda_2$ and $Ax=\lambda_1x$, $Ay=\lambda_2y$. Then $$(Ax,y)-(x,Ay)=(\lambda_1x,y)-(x,\lambda_2y)=(\lambda_1-\lambda_2)(x,y)$$ But the left side is $0$, because of the symmetry :$(Ax,y)=Ax\cdot y=x\cdot A^Ty=x\cdot Ay=(x,Ay)$. Since $\lambda_1-\lambda_2\neq 0\Rightarrow (x,y)=0$