# negative semidefinite matrices

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.

• 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 $L$? Dec 23, 2022 at 13:52
• @KwinvanderVeen $Q$ and $F$ are real. You are right, it should've been $LL^T$. Dec 23, 2022 at 13:56

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 $$\star$$.

• 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? Dec 23, 2022 at 17:43
• Also following your first equation, shouldn't the RHS of $\circ$ be positive? Dec 23, 2022 at 17:58
• @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. Dec 23, 2022 at 18:56
• @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. Dec 23, 2022 at 19:14
• @Morcus I have updated my answer. Is it now clear or do you still have questions? Dec 24, 2022 at 12:49