Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have $n \times n$ real matrices $A$ and $D$. $D$ is diagonal. Let's $v_i(A), \lambda_i(A)$ be a couple of eigenvectors-eigenvalues of $A$. What relationships there exists between $v_i(B), \lambda_i(B)$ and $v_i(A), \lambda_i(A)$ where $B = DA$?

share|cite|improve this question
up vote 5 down vote accepted

Left-multiplication be a diagonal matrix does not have any simple effect on eigenvalues, and given that eigenvalues are perturbed (or destroyed) what could one possibly want to say about "corresponding" eigenvectors?

Example in $\def\R{\Bbb R}\R^2$. With $A=(\begin{smallmatrix}0&1\\1&0\end{smallmatrix})$ one has eigenvalues $+1,-1$ with eigenvectors that you can easily spot. Multiplying by $D=(\begin{smallmatrix}1&0\\0&-1\end{smallmatrix})$ one gets $B=DA=(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix})$, without any real eigenvalues at all. ($B$ has complex eigenvalues $\pm\mathbf i$, with eigenvectors that bear no relation to those of $A$).

share|cite|improve this answer
One could define $f(x) = (1-x)A + xB$ for $x \in [0, 1]$; as this function is continuous, its eigenvectors vary continuously, and the eigenvectors corresponding to simple eigenvalues also vary continuously. So at least for simple eigenvalues, it makes sense to talk about the "same eigenvectors". – Solomonoff's Secret Oct 1 '14 at 15:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.