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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} $.

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  • $\begingroup$ I don't see where you assume that $A$ is diagonalizable. $\endgroup$
    – Zuy
    Commented Aug 11, 2020 at 9:48
  • $\begingroup$ If I do not assume $A$ is diagonalizable, how do we know that $f$ splits? $\endgroup$
    – shk910
    Commented Aug 11, 2020 at 10:20
  • $\begingroup$ You don't need this. Let me write an answer. $\endgroup$
    – Zuy
    Commented Aug 11, 2020 at 10:24

1 Answer 1

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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.

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