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Let $A \in M_n(\mathbb C)$ and $\lambda$ is an eigenvalue. Does there exist a sequence of diagonalizable matrices $D_n$ and a sequence $\{\lambda_n\}$ complex numbers such that each $\lambda_n$ is an eigenvalue of $D_n$, $D_n \to A$ and $\lambda_n \to \lambda$ ?

I know that the set of diagonalizable matrices are dense in $M_n(\mathbb C)$ , but I'm not sure whether that implies that given a matrix and an eigenvalue of it, whether we can approximate the eigenvalue via eigenvalues of some sequence diagonalizable matrices which approximates the matrix.

Please help.

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Let $p_i(x)$ and $p(x)$ be the characteristic polynomials of $D_i$ and $A$. Since $D_i \rightarrow A$, we have $p_i \rightarrow p$ as $i \rightarrow \infty$. Since $p(\lambda)=0$, we have $p_i(\lambda) \rightarrow 0$. Write $p_i(x)=(x-\mu_{i1})\cdots (x-\mu_{in})$. So $$\|\lambda-\mu_{i1}\|\cdots \|\lambda-\mu_{in}\| \rightarrow 0.$$ This implies that $\min_{t}\|\lambda-\mu_{it}\|$ converges to zero as $i\rightarrow \infty$.

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In general: if $D_m \to A$, then for any $\lambda \in A$ we can necessarily find sequence $\lambda_m \to \lambda$ where $\lambda_m$ is an eigenvalue of $\lambda$.

That being said: for your weaker statement that there exists such a $D_m$ and $\lambda_m$ with $D_m \to A$, a construction like this one will suffice. In particular: if $A = SJS^{-1}$ (where $J$ is in Jordan normal form), then we can define $ D_m = S[K_mJ]S^{-1}$, where we take $K_m$ to be the diagonal matrix $$ K_m = \pmatrix{1 - \frac 1m \\ & \ddots \\ && 1 - \frac nm} $$ Since $K_m J$ is upper triangular, it's easy to see that its eigenvalues converge to those of $J$. Moreover, $K_m J$ will have distinct eigenvalues (and will therefore be diagonalizable) for all but finitely many values of $m$.

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  • $\begingroup$ Note: I have used $m$ as the index of the sequence to distinguish it from $n$, the size of $A$. $\endgroup$ – Omnomnomnom Sep 24 '18 at 2:24

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