# Eigenvalues and multiplication by diagonal matrices

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 $B = DA$?

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## 1 Answer

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 $A$).

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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