Let $T$ be a linear operator on a finite-dimensional vector space $V$
with the distinct eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_k$ and
corresponding multiplicities $m_1,m_2,\dots,m_k$ Suppose that $\beta$ is
a basis for $V$ such that $[T]_\beta$ is an upper triangular matrix.
Prove that the diagonal entries of $[T]_\beta$ are
$\lambda_1,\lambda_2,\dots,\lambda_k$ and that each $\lambda_i$ occurs
$m_i$ times $(1 \le i \le k)$.
Proof :Let $T$ be a linear map on a finite-dimensional vector space $V$ with distinct eigenvalues $\lambda_1,\dots,\lambda_k$ and with their corresponding multiplicities $m_1,\dots,m_k$. Since $[T]_\beta$ is an upper triangular matrix, and from the conditions given, we have $f(\lambda)=\Pi_{i=1}^k(\lambda_i-\lambda)^{m_i}$. Hence proved. Can I leave it there?
Best Answer
Your argument is correct if you have shown before the equation for $f(\lambda)$. If you want to extend it a little bit, you can do it the following way:
The characteristic polynomial of $T$ is independent of the choice of $\beta$. Hence, the polynomial
$$f(t) = \det ( [T]_\beta - t I) = \prod_{i = 1}^n (([T]_\beta)_{ii} - t)$$
splits into linear factors, where the second equality is true since $T$ is upper triangular. But then it's diagonal entries are the zeroes of $f(t)$.