Suppose $D$ is $n$ by $n$ (real) diagonal matrix (If necessary, assume every $D_{ii}$ is strictly positive) and $A$ is arbitrary (real) diagonalizable $n$ by $n$ matrix. In other words, eigenvectors of $A$ spans $\mathbb{R}^n$.

Then, is there any explicit relationship between eigenvalue/eigenvectors of $DA$ and $A$? For example, if we know eigenvectors of $A$, is there any way to construct eigenvectors of $DA$?

Since $D$ is diagonal matrix, which is very simple, maybe there is some way to connect $DA$ and $A$, but I couldn't figure it out. Is there any explicit relationship?


In general, there is even no reason for $DA$ to have any eigenvectors. Example: $$\left(\matrix{1&0\\0&-1}\right)\underbrace{\left(\matrix{0&-1\\-1&0}\right)}_{\text{eigenvalues $\pm 1$}}=\underbrace{\left(\matrix{0&-1\\1&0}\right)}_{\text{no real eigenvalues}}$$

The reason you should not be expecting anything simple is that the reasons that make you feel that $A$ and $D$ should be easy enough to deal with live in different worlds: $A$ is diagonalisable, which means "nice" in a certain basis; $D$ is diagonal, which means "nice" in the current basis. You can't have both these nice properties at once.

Just for the sake of completeness, having the entries of $D$ positive does not help:

$$\left(\matrix{0.1&0\\0&1}\right)\underbrace{\left(\matrix{10&-10\\2&1}\right)}_{\text{eigenvalues $5$ and $6$}}=\underbrace{\left(\matrix{1&-1\\2&1}\right)}_{\text{no real eigenvalues}}$$

  • $\begingroup$ Thank you for the simple explanation. It's not that simple, then... $\endgroup$ – KGEO Jun 11 '18 at 20:08
  • $\begingroup$ @SunTaek You're welcome. No indeed. $\endgroup$ – Arnaud Mortier Jun 11 '18 at 20:12

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