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

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


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

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 ?


Here is a shorter solution: You need only the symmetry of the matrix.

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 $\lambda_1-\lambda_2\neq 0\Rightarrow (x,y)=0$


Yes, eigenvectors belonging to distinct eigenvalues of symmetric positive matrix are orthogonal, and your solution is correct.

More generally, this follows from the fact, that symmetric positive matrices are hermitian, and therefore normal (which is the most general class of matrices having this property).


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