Upper triangular matrix and diagonalizability

Let $$A$$ be an $$n \times n$$ matrix that is similar to an upper triangular matrix and has the distinct eigenvalues $$\lambda_1, \lambda_2, ... , \lambda_k$$ with corresponding multiplicities $$m_1, m_2, ... , m_k$$. Prove the following statements.

(a) $$\operatorname{tr}(A) = \sum_{i=1}^k m_i \lambda_i$$

(b) $$\det(A) = (\lambda_1)^{m_1} (\lambda_2)^{m_2} \cdots(\lambda_k)^{m_k}.$$

I am wondering if we can solve this question without assuming that $$A$$ is diagonalizable. With this assumption, the characteristic polynomial $$f(t)$$ of $$A$$ splits, so $$f(t) = (\lambda_1 - t)^{m_1}\cdots (\lambda_k - t)^{m_k}$$. Since $$\det(A - t I) = \det(D - tI)$$, where $$D$$ is an upper triangular matrix with $$D = Q^{-1} A Q$$ for some invertible matrix $$Q$$. Since $$D$$ is an upper triangular matrix, $$\det(D -tI) = (D_{11} -t) \cdots (D_{nn} - t)$$. Thus, $$(\lambda_1 - t)^{m_1}\cdots (\lambda_k - t)^{m_k} = (D_{11} -t) \cdots (D_{nn} - t)$$. Noting that $$\operatorname{tr}(A) = \operatorname{tr}(D)$$, $$\operatorname{tr}(A) = \sum_{i=1}^n D_{ii} = \sum_{i=1}^k m_i \lambda_i$$. Similarly, $$\det(A) = \det(D) = \prod_i(D_ii) =(\lambda_1)^{m_1} (\lambda_2)^{m_2} \cdots(\lambda_k)^{m_k}$$.

• I don't see where you assume that $A$ is diagonalizable.
– Zuy
Commented Aug 11, 2020 at 9:48
• If I do not assume $A$ is diagonalizable, how do we know that $f$ splits? Commented Aug 11, 2020 at 10:20
• You don't need this. Let me write an answer.
– Zuy
Commented Aug 11, 2020 at 10:24

You can use the following well-known

Properties

Let $$A$$ and $$B$$ be $$n\times n$$-matrices such that $$B$$ is invertible. Then $$\det (B^{-1} A B)=\det A$$

and

$$\mathrm{Tr}(B^{-1} A B)=\mathrm{Tr}A.$$

If we let $$M$$ be the upper triangular matrix you mentioned in your question, and $$B$$ the invertible $$n\times n$$-matrix satisfying $$B^{-1}MB=A,$$ then $$\mathrm{Tr}A=\mathrm{Tr}(B^{-1}MB)=\mathrm{Tr}M=\sum_{i=1}^k m_i\lambda_i$$ and similarly $$\det A=\det(B^{-1}MB)=\det M=\lambda_1^{m_1}\cdots\lambda_k^{m_k}.$$

Note that the first equalities are by equality of $$A$$ and $$B^{-1}MB$$, the second equalities follow from above properties, and the third equalities hold because $$M$$ is upper triangular.