# Can we approximate any eigenvalue of a matrix via eigenvalues of some sequence diagonalizable matrices which approximates the matrix?

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.

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