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We have a $n\times n$ matrix $A$ , which is also block diagonalized into $$ \left(\begin{matrix} A_1 & & &\\ & A_2 & &\\ & & \ddots & \\ & & & A_s \end{matrix}\right) $$ And we also know that A can be diagonalized, how to prove that each block $A_i$ can be diagonalized?

I've already known that this can be proved using Jordan canonical form, but I wonder if there is another way to prove this using more fundamental knowledge (Suppose I've leant no more than the eigenvalues and eigenvectors of matrices)

Below are my progress:

Suppose A has distinct eigenvalues $\lambda_1,\lambda_2,\cdots,\lambda_s$, then we have $ A(\vec{\beta_1^{(1)}},\cdots,\vec{\beta_{a_1}^{(1)}},\cdots,\vec{\beta_1^{(s)}},\cdots,\vec{\beta_{a_s}^{(s)}})=$ $$(\vec{\beta_1^{(1)}},\cdots,\vec{\beta_{a_1}^{(1)}},\cdots,\vec{\beta_1^{(s)}},\cdots,\vec{\beta_{a_s}^{(s)}}) \left(\begin{matrix} \lambda_1 & & & & & & \\ &\ddots & & & & &\\ & & \lambda_1 & & & &\\ & & & \ddots & & & \\ & & & & \lambda_s & &\\ & & & & & \ddots & \\ &&&&&&\lambda_s \end{matrix}\right) $$

if I could prove that each block $A_i$'s size is also $a_i\times a_i$ (don't use Jordan canonical form!), then my problem can be easily solved. However, this is where I'm stuck. Any help is appreciated!

Rodri
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    Have you learned about minimal polynomials, invariant (or stable) subspaces, and the theorem that a linear operator (corresponds to a square matrix) is diagonalizable if and only if its minimal operator is the product of distinct linear factors? These would give a simple proof to the result. – Duong Ngo Dec 30 '24 at 07:38
  • @Duong Ngo Well, what I want is to use the basic knowledge to solve this... Your answer is a good way to comprehend this question, but I need a more rigorous proof – Rodri Dec 30 '24 at 08:15
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    @Rodri Answers here would help. – Duong Ngo Dec 30 '24 at 08:20
  • Please do not answer in the comments, @ToniMhax. – Shaun Dec 30 '24 at 18:31

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(This is a deleted comment i wrote). The block diagonalized form of $A$ call it $B$ is diagonalizable since $A$ is diagonalizable. Suppose one $A_i$ is not diagonalizable, this means that one eigenvalue $\lambda_i$ of $A_i$ which is also of $B$ has a geometric multiplicity less than its algebraic multiplicity. But this is a contradiction since the eigenvectors of $\lambda_i$ as an eigenvalue for $B$ are the eigenvectors of $A_i$ (for $\lambda_i$) completed by zero entries.

Toni Mhax
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