0
$\begingroup$

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} $.

$\endgroup$
3
  • $\begingroup$ I don't see where you assume that $A$ is diagonalizable. $\endgroup$ – Zuy Aug 11 '20 at 9:48
  • $\begingroup$ If I do not assume $A$ is diagonalizable, how do we know that $f$ splits? $\endgroup$ – shk910 Aug 11 '20 at 10:20
  • $\begingroup$ You don't need this. Let me write an answer. $\endgroup$ – Zuy Aug 11 '20 at 10:24
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.