Can someone point me to a paper, or show here, why symmetric matrices have orthogonal eigenvectors? In particular, I'd like to see proof that for a symmetric matrix $A$ there exists decomposition $A = Q\Lambda Q^{-1} = Q\Lambda Q^{T}$ where $\Lambda$ is diagonal.

  • $\begingroup$ Are $Q$ and $F$ real or complex matrices (and in the complex case shouldn't the transposes be a Hermitian transpose?)? And in your first equation are all sings correct and should there maybe be a (Hermitian) transpose on one of the two $L$? $\endgroup$ Dec 23, 2022 at 13:52
  • $\begingroup$ @KwinvanderVeen $Q$ and $F$ are real. You are right, it should've been $LL^T$. $\endgroup$
    – Morcus
    Dec 23, 2022 at 13:56

1 Answer 1


Since the matrix $P$ is only required to be symmetric for $\star$ and the statement you want to show and nonnegative definite is the same a positive semi-definite. So using $P'=-P$ would still qualify as symmetric, so substituting this into $\star$ yields

$$ Q^\top (P'\,F + F^\top P')\,Q = -Q^\top (P\,F + F^\top P)\,Q \preceq 0, \tag{$\times$} $$

with $A \preceq 0$ indicating that $A$ is negative semi-definite. Note that this does not mean that $P'\,F + F^\top P' \preceq 0$. Namely, it could be that that resulting matrix is indefinite, however the positive semi-definite "contribution" to that matrix would lie in the null space/Kernel of $Q$. Namely, in general an indefinite symmetric real matrix $A\in\mathbb{R}^{n\times n}$ has the following eigendecomposition

$$ A = \begin{bmatrix} V & U \end{bmatrix} \begin{bmatrix} \Omega & 0 \\ 0 & \Phi \end{bmatrix} \begin{bmatrix} V^\top \\ U^\top \end{bmatrix} = V\,\Omega\,V^\top + U\,\Phi\,U^\top, $$

with $\Omega\in\mathbb{R}^{m\times m}\preceq0$ and $\Phi\in\mathbb{R}^{k\times k}\succeq0$ diagonal matrices, $k+m=n$, $V\in\mathbb{R}^{n\times m}$ and $U\in\mathbb{R}^{n\times k}$, such that $\begin{bmatrix}V&U\end{bmatrix}$ is a orthonormal matrix. Note that $X=V\,\Omega\,V^\top\preceq0$ and $Y=U\,\Phi\,U^\top\succeq0$. However, in order for $\times$ to be true and because the eigenvectors are orthogonal it should hold that $Q^\top Y\,Q =0$. Such that the LHS of $\times$ can also be written as

$$ Q^\top (P'\,F + F^\top P')\,Q = Q^\top X\,Q\preceq 0, $$

since if $Q^\top Y\,Q \neq0$ then $\times$ would also be indefinite.

The difference between $\star$ and your statement is the additional term $Q^\top L\,L^\top Q$. That additional term can be shown to be positive semi-definite. And thus there should exists a $L$ such that

$$ Q^\top L\,L^\top Q = -Q^\top (P'\,F + F^\top P')\,Q. \tag{$\circ$} $$

One such solution could be found by solving $L\,L^\top = -X \succeq 0$, with $X$ as defined previously; for example using the Cholesky decomposition. Moving all terms of $\circ$ to one side and factoring out $Q$ yields your statement

$$ Q^\top (P'\,F + F^\top P' + L\,L^\top)\,Q = 0, $$

thus showing that it is equivalent to $\star$.

  • $\begingroup$ Thanks a lot for the clarification! I don't follow clearly why $X \preceq 0$, $Y \succeq 0$ and $Q^\top Y\,Q=0$ should hold. Any hint? $\endgroup$
    – Morcus
    Dec 23, 2022 at 17:43
  • $\begingroup$ Also following your first equation, shouldn't the RHS of $\circ$ be positive? $\endgroup$
    – Morcus
    Dec 23, 2022 at 17:58
  • $\begingroup$ @Morcus The RHS of $\circ$ has a negative sign in front. So using the first equation from my answer and negating it yields that the RHS of $\circ$ is positive semi-definite. $\endgroup$ Dec 23, 2022 at 18:56
  • $\begingroup$ @Morcus Regarding your first comment, I only introduced $X$ and $Y$ to hopefully clarify without having to add too much details. But note that in general adding a positive semi-definite and a negative semi-definite matrix does not guarantee that its sum is indefinite. I will add a more detailed explanation using the eigendecomposition to my answer about why this can be done. $\endgroup$ Dec 23, 2022 at 19:14
  • $\begingroup$ @Morcus I have updated my answer. Is it now clear or do you still have questions? $\endgroup$ Dec 24, 2022 at 12:49

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