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I finished a whole chapter about eigenvalues, eigenvectors and subspaces which are invariant (w.r.t. a linear operator i.e. a mapping from $V$ to $V$).

I went through lots of theorems and propositions and I understand their proofs well. But I think a question came up which is not covered there. Here it is.

Let's say $V$ is $n$-dimensional vector space over $\mathbb{R}$ ($n$ - natural number). Also, let's say $\phi : V \to V$ is a linear map. Let's say the characteristic polynomial of $\phi$ has real roots $\lambda_1, \lambda_2, \dots, \lambda_k$ with multiplicies $m_1, m_2, \dots, m_k$

Of course this implies

$\sum_1^{k}m_k \le n$

Then (for any $i$), is it true that in $V$ we can find $m_i$ linearly independent eigenvectors corresponding to $\lambda_i$, but we cannot find $m_i+1$ linearly independent eigenvectors corresponding to $\lambda_i$?

I solved a few problems (with actual matrices containing numbers), and that always seems to be the case.

E.g. if the characteristic polynomial is divisible by $(x-5)^2$, then I am finding exactly 2 linearly independent eigenvectors corresponding to the eigenvalue 5 (that's because the multiplicity is 2). So basically in that case there's a 2-dimensional $\phi$ invariant subspace of $V$ corresponding to the eigenvalue 5:

$U = \{u : \phi(u) = 5u\}$

$\dim\ U = 2$

But is this always the case? And can $U$ happen to be higher dimensional than $2$ ($2$ being just an example - the multiplicity of the eigenvalue $5$, when viewed as a characteristic root of $\phi$)?

I don't think any of the theorems in that chapter covered these two questions.

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  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ Commented Apr 10, 2022 at 19:20
  • $\begingroup$ If $\phi$ is not diagonalizable then you do not have equality between the multiplicity of one of the $\lambda_i$ and the dimension of the corresponding eigenspace. If you want simple counter example for any order, then you can choose a triangular matrix with a constant diagonal and non zero coefficients on the top of the main diagonal, it will never be a diagonalizable matrix. $\endgroup$ Commented Apr 10, 2022 at 19:33

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The dimension of the eigenspace corresponding to the eigenvalue $\lambda_i$ is smaller than or equal to $m_i$, and therefore you cannot possibly find $m_i+1$ find $m_i+1$ linearly independent eigenvectors corresponding to $\lambda_i$.

On the other hand, the dimension can be smaller than $m_i$. Take, for instance $n=2$ and $\phi(x,y)=(y,0)$. The characteristic polynomial is $\lambda^2$, but the dimension of the eigenspace corresponding to the eigenvalue $0$ is $1$, rather than $2$.

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  • $\begingroup$ Hm... Interesing... Thank you. So $\dim\ U \le m_i$, not always $=m_i$. OK, I will verify the example you're giving me. $\endgroup$ Commented Apr 10, 2022 at 19:21
  • $\begingroup$ Do you have a higher order example? E.g. when V is 3-dimensional and I get 2 roots with multiplicities 2 and 1. But $\dim\ U = 1$ (for the first i.e. the 2-fold root). $\endgroup$ Commented Apr 10, 2022 at 19:23
  • $\begingroup$ Sure: $\phi(x,y,z)=(y,0,z)$. Then the characteristic polynomial is $\lambda^2(1-\lambda)$. $\endgroup$ Commented Apr 10, 2022 at 19:24
  • $\begingroup$ Great, thanks a lot. $\endgroup$ Commented Apr 10, 2022 at 19:24

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