# Eigenvalues with different algebraic multiplicities of an upper-triangular matrix

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?

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 $$f(t)$$.