Necessary and sufficient condition to have an upper-triangular matrix

Question

The following is result 5.44 from Linear Algebra Done Right, fourth edition by Sheldon Axler:

Theorem. Suppose $V$ is finite-dimensional and $T \in \mathcal{L}(V)$. Then $T$ has an upper-triangular matrix with respect to some basis of $V$ if and only if the minimal polynomial of $T$ equals $(z-\lambda_1) \cdots (z-\lambda_m)$ for some $\lambda_1, \ldots, \lambda_m \in \mathbf{F}$.

Proof of forward direction. First suppose $T$ has an upper-triangular matrix with respect to some basis of $V$. Let $\alpha_1, \ldots, \alpha_n$ denote the diagonal entries of that matrix. Define a polynomial $q \in \mathcal{P}(\mathbf{F})$ by

$$ q(z) = (z - \alpha_1) \cdots (z - \alpha_n) $$

Then $q(T)=0$, by 5.40. Hence $q$ is a polynomial multiple of the minimal polynomial of $T$, by 5.29. Thus the minimal polynomial of $T$ equals $(z - \lambda_1) \cdots (z - \lambda_m)$ for some $\lambda_1, \ldots, \lambda_m \in \mathbf{F}$ with $\{\lambda_1, \ldots, \lambda_m\} \subseteq \{\alpha_1, \ldots, \alpha_n\}$.

Something seems off to me. We're letting $\alpha_1, \ldots, \alpha_n$ denote the diagonal entries of the upper-triangular matrix of $T$. By result 5.41 in the book, this means that $\alpha_1, \ldots, \alpha_n$ are the eigenvalues of $T$. By result 5.27(a), these must be the zeros of the minimal polynomial. But then in the proof, we say that $\{\lambda_1, \ldots, \lambda_m\} \subseteq \{\alpha_1, \ldots, \alpha_n\}$, which suggests to me there may be an $\alpha \in \{\alpha_1, \ldots, \alpha_n\}$ that is not in $\{\lambda_1, \ldots, \lambda_m\}$. If that's the case, then how can $(z - \lambda_1) \cdots (z - \lambda_m)$ possibly be the minimal polynomial if it is missing some $\alpha$ as a zero?

Answer 1

That's not what happens; Axler should really be talking about multisets here rather than sets. The issue has to do with multiplicities. For example, if

$$A = \begin{bmatrix} \lambda & 1 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix}$$

then the minimal polynomial is $(t - \lambda)^2$ but the polynomial $q$ constructed by taking the product of the diagonal entries (which is the characteristic polynomial) is $(t - \lambda)^3$. So they have the same roots (they have to, because the roots have to be the eigenvalues of $A$) but the minimal polynomial has those roots with smaller multiplicity.

Generally, assuming that we're working over an algebraically closed field for simplicity, the multiplicity of an eigenvalue $\lambda$ of $A$ as a root of the characteristic polynomial is called its algebraic multiplicity. It must also be a root of the minimal polynomial $m$, but with multiplicity $\le$ the algebraic multiplicity. The exact multiplicity can be described using the Jordan normal form of $A$: it's the maximum size of a Jordan block with eigenvalue $\lambda$ (which in the above example is $2$), whereas the algebraic multiplicity is the sum of the sizes of all the Jordan blocks with eigenvalue $\lambda$. As a simple example, $A$ could be a diagonal matrix with entries $\lambda$, and then the minimal polynomial would be $t - \lambda$ but the characteristic polynomial would be $(t - \lambda)^n$.