If $A$ is a matrix with negative eigenvalues, then $\exists M$ : $A = -MM^T$

Question

Let $A$ be a symmetric matrix with all its eigenvalues negative. Prove that there exists a matrix $M$ such that : $A = -MM^T$.

Now, regarding my question, I have found another older question, that though goes the "opposite" way with non-negative eigenvalues : Non-negative Eigenvalues.

Now what I know is that : $Av = -λv$ for our negative eigenvalues. Also, $A$ is symmetric, which means that $AA^T = I$. We can then form the equation $Av + λv = AA^T - I$ which maybe could lead somewhere, but I cannot figure out how to continue from then on.

Answer 1

If $A$ is symmetric and negative definite (i.e., all its eigenvalues are negative), then $-A$ is symmetric and positive definite (i.e., all its eigenvalues are positive).

From the symmetry of $-A$, we conclude that it is diagonalizable, that its eigenvalues are real, and that its eigenvectors are orthogonal. Hence, $-A$ has an eigendecomposition $-A = Q \Lambda Q^T$, where the columns of $Q$ are the eigenvectors of $-A$ and the diagonal entries of diagonal matrix $\Lambda$ are the eigenvalues of $-A$. From the positive definiteness of $-A$, we conclude that we can take the square roots of its eigenvalues. Hence,

$$-A = Q \Lambda Q^T = Q \Lambda^{\frac 12} \Lambda^{\frac 12} Q^T = (Q \Lambda^{\frac 12}) (Q \Lambda^{\frac 12})^T$$

Let $M := Q \Lambda^{\frac 12}$. Thus, $-A = M M^T$, or, $A = - M M^T$. Note that the columns of $M$ are the eigenvectors of $-A$ multiplied by the square roots of the eigenvalues of $-A$.

Answer 2

Easily we see that $-A$ is symmetric with positive eigenvalues and by spectral theorem it's diagonalizable in orthonormal basis: there is an orthogonal matrix $P$ and $\lambda_1,\ldots,\lambda_n>0$ such that

$$-A=P DP^T$$ where $D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$. Let $\Delta=\operatorname{diag}(\sqrt\lambda_1,\ldots,\sqrt\lambda_n)$ then $\Delta^2=D$ and with $M=P\Delta$ we have the desired result.