Eigenvalues of the Product of a Diagonal and a Symmetric Matrix

Question

Let $A \in \mathbb{R}^{n\times n}$ be a symmetric matrix and $D \in \mathbb{R}^{n\times n}$ be a diagonal matrix with positive entries. Prove that the matrix $P:=DA$ has real eigenvalues.

Answer 1

The characteristic equation of matrix $P$ is \begin{align} \text{det}(\lambda I - DA) = \text{det}( D^{\frac{1}{2}} (\lambda I - D^{\frac{1}{2}}AD^{\frac{1}{2}}) D^{\frac{-1}{2}}) =0 \end{align} Thus the eigenvalues of $P=DA$ are the same as the eigenvalues of $Q:=D^{\frac{1}{2}}AD^{\frac{1}{2}}$ which is symmetric and has real eigenvalues.

Answer 2

In general, two matrices having the same characteristic polynomial may not be similar, therefore they may not have the same eigenvalues. More specifically, these two matrices must be non-derogatory. See page 660 of Meyer (2000):

If A and B are non-derogatory matrices that have the same characteristic polynomial, then A is similar to B.

By modifying @Arthur's solution, we just use the definition of similar matrices: $M$ and $N$ are said to be similar whenever there exists a nonsingular matrix $R$ such that $R^{−1}MR = N$.

In this case, we take nonsingular matrix $R = D ^ {- 1/2}$ and $M = D^{1/2}AD^{1/2}$, then $R^{−1}MR = D^{1/2}(D^{1/2}AD^{1/2})D ^ {- 1/2} = DA$. Therefore, $D^{1/2}AD^{1/2}$ and $DA$ are similar by definition, then they have the same eigenvalues. However, $D^{1/2}AD^{1/2}$ is real symmetric, so it must have real eigenvalues only. So, $DA$ has real eigenvalues only.

[1] Meyer, Carl D., and Ian Stewart. Matrix analysis and applied linear algebra. Society for Industrial and Applied Mathematics, 2000.