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Let $A \in \Bbb C^n$ and suppose we have already established that $\Bbb C^n$ may be decomposed in the direct sum of generalized eigenspaces of $A$ (the generalized eigenspace of $A$ relative to the eigenvalue $\lambda_j$ being defined as $\text{ker}(A-\lambda_j I)^{\nu_j}$ where $\nu_j$ denotes the algebraic multiplicity of $\nu_j$). Is there any easy way of proving that the dimension of the generalized eigenspace of $A$ relative to $\lambda_j$ is exactly $\nu_j$ once we have established this decomposition? It seems to me that it shouldn’t be too much difficult but I couldn’t think of anything.

Matteo Menghini
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  • Here are a couple of approaches: (1) carefully go through your proof(s) that $\mathbb C^n$ is a direct sum of the generalized eigenspaces of $A$ as they may already contain / imply the result. E.g. see my proof here https://math.stackexchange.com/questions/3878092/prove-that-the-intersection-of-2-generalised-eigenspaces-is-the-zero-space/ . – user8675309 Jul 28 '23 at 16:06
  • (2) This will take some work by you: let $\mathbf B_1, \mathbf B_2$ be bases for the kernel and image of $(A-\lambda_j I)^n$ respectively and $\mathbf B:=\bigg[\begin{array}{c|c|c|c} \mathbf B_1 & \mathbf B_2 \end{array}\bigg]$ Then $A\mathbf B = \mathbf B\left[ \begin{matrix}A_1 & \mathbf 0\ \mathbf 0 & A_2\end{matrix}\right]$,so what could the characteristic polynomial be for $A_1$ given that $A_1$ only has $\lambda_j$ as eigenvalue and $A_2$ doesn't have $\lambda_j$ as an eigenvalue. – user8675309 Jul 28 '23 at 16:14
  • @user8675309 I am sorry but I couldn’t understand the equality you claim. I have tried reconstructing $(A-\lambda_j)^{\nu_j}$ in order to obtain it but to no avail. Also, did you mean $\nu_j$ instead of $n$? Anyway, thanks for the first comment! – Matteo Menghini Jul 28 '23 at 22:25
  • for my second comment: no... you have $C:=\big(A-\lambda_j I\big)$ where $C\in \mathbb C^{n\times n}$ so by basic rank property manipulation (i.e. a much more basic notion that you should encounter long before generalized eigenspaces) you have $\mathbb C^n = \text{image }C^n \oplus \ker C^n$. What you are trying to prove, in some sense is that $\text{rank} C^{v_j} \geq \text{rank}C^{n}$ is always met with equality but you don't "know" this yet so I use $n$ as an exponent. – user8675309 Jul 29 '23 at 00:23
  • Are you familiar with the Jordan normal form? I suspect showing the above is equivalent to finding a basis for $\ker (A-\lambda_j I)^{\nu_j}$ that results in the JNF. The generalised eigenvectors are the connection between the characteristic polynomial and the dimension of the above space. – copper.hat Jul 29 '23 at 18:24

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