Axler's LADR. Algebraic multiplicity of an eigenvalue is the number of times it appears on the diagonal of an upper triangular matrix?

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.

• +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 $V$. Feb 15, 2018 at 14:27
• 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?
– user491874
Feb 15, 2018 at 14:47
• @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.
– Anu
Feb 15, 2018 at 14:51
• @Anu I see, thanks!
– user491874
Feb 15, 2018 at 15:14