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This problem is from Sheldon Axler's Linear Algebra Done Right, Chapter 8. Let $T \in \mathcal{L}(V)$, where $V$ is a finite dimensional complex vector space. If the matrix of $T$ is upper triangular with respect to any basis of $V$, the number of times $\lambda$ appears on the diagonal of this matrix equals the (algebraic) multiplicity of $\lambda$ as an eigenvalue of $T$.

It suffices to show that $\dim \text{null } T^n = \dim G(0, T)$ equals the number of $0$'s on the diagonal, where $G(0, T)$ is the space of generalized eigenvectors of $T$ with respect to $0$.

I found this rather nice proof on this blog, which uses induction on the dimension of $V$.

I'm interested in finding alternative ways to prove this statement, which may provide a different way of looking at it.

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    $\begingroup$ +1, this was the unique exercise in the whole book that I couldnt solve. I found the same proof using induction over the dimension of $V$. $\endgroup$ – Masacroso Feb 15 '18 at 14:27
  • $\begingroup$ I am puzzled, if you take $\det(T-\lambda I_n)$, you will see the factor $(\lambda-\alpha)$ in the result exactly as many times $\alpha$ appears on the diagonal - simply because the determinant of a triangular matrix is the product of the entries on the diagonal. Where is the catch? In other words, why even talk about generalised eigenvectors? $\endgroup$ – user491874 Feb 15 '18 at 14:47
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    $\begingroup$ @user8734617 Axler's book defines the multiplicity of $\lambda$ as the dimension of $G(\lambda, T)$ and the characteristic polynomial as $(z - \lambda_1)^{d_1}\cdots(z - \lambda_m)^{d_m}$ where $\lambda_1,\dots,\lambda_m$ are the distinct eigenvalues of $T$, and $d_1,\dots,d_m$ their corresponding multiplicities. It's only later in the book that he shows that the more familiar definition of the characteristic polynomial involving determinants is equivalent to this one. $\endgroup$ – Anu Feb 15 '18 at 14:51
  • $\begingroup$ @Anu I see, thanks! $\endgroup$ – user491874 Feb 15 '18 at 15:14

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