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For a non-diagonalizable $n \times n$ matrix, how to prove that $\exists (A_m)$ is diagonalizable matrix such that $A_m$ converges to $A$. Here we define norm on $M(n, \Bbb C)$ s.t $||X||=(\sum _{i=1}^\infty\sum_{j=1}^\infty|x_{ij}|^2)^{1/2}$ and $A_m$ converges to $A$ iff $||A_m - A|| \to 0$ as $m\to \infty$

I was thinking to use Jordon Canonical form but can't get it as the calculation are getting twisted. If someone has some detailed proof please help.

José Carlos Santos
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Ri-Li
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  • see answer here: https://math.stackexchange.com/questions/107945/diagonalizable-matrices-with-complex-values-are-dense-in-set-of-n-times-n-comp – Mathematician 42 Jan 25 '18 at 14:29

2 Answers2

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The set of all diagonalizable matrices is dense in the space of all $n\times n$ matrices; you'll find a proof here. This is enough to prove what you want.

José Carlos Santos
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  • This is the exact claim the OP wants to prove, so certainly enough. – Pedro Jan 25 '18 at 14:34
  • @PedroTamaroff No. It is equivalent, but it is not the same statement. However, if the OP states that, for him or her, the statements are trivially equivalent, then I will delete my answer. – José Carlos Santos Jan 25 '18 at 14:36
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As you remark it boils down to Jordan normal normal form, from which we see it suffices to handle the nilpotent case. For which we have for example

$$\begin{pmatrix} \epsilon&1&0\\ 0&\eta&1\\ 0&0&\nu \end{pmatrix}\to \begin{pmatrix} 0&1&0\\ 0&0&1\\ 0&0&0 \end{pmatrix}$$

Where $\epsilon, \eta, \nu$ limit to zero in such a way that they are distinct, and so the matrix is diagonalizable, its minimal polynomial being the product of distinct factors.

Rene Schipperus
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