Show that $\det(A)=(\lambda_1)^{m_1}(\lambda_2)^{m_2}…(\lambda_k)^{m_k}=\prod_{j=1}^k(\lambda_i)^{m_j}$

Question

This is the problem... I got part a and part b... but I'm lost for part c.
Anyone have a hint of something I can try?

Answer 1

By assumption there exists an invertible matrix $S$ such that $SAS^{-1}=D$ is a diagonal matrix, i.e., $D=\rm {diag}(\lambda_1^{m_1},\cdots ,\lambda_k^{m_k})$. Then the claim is obvious, since $$ \det(D)=\det(SAS^{-1})=\det(S)\det(A)\det(S^{-1})=\det(A). $$

Answer 2

Since $\;A\;$ is diagonalizable there exists invertible $\;P\;$ such that

$$A=P^{-1}\begin{pmatrix}\lambda_1&0&\ldots&\ldots&0\\\ldots&\ldots&\ldots&\ldots&\ldots\\ 0&\ldots&\lambda_1&\ldots&0\\ \ldots&\ldots&\ldots&\ldots&\ldots\\ 0&\ldots&0&\lambda_k\ldots&0\\ \ldots&\ldots&\ldots&\ldots&\lambda_k\end{pmatrix}P\implies$$

$$ \det A=\det\begin{pmatrix}\lambda_1&0&\ldots&\ldots&0\\\ldots&\ldots&\ldots&\ldots&\ldots\\ 0&\ldots&\lambda_1&\ldots&0\\ \ldots&\ldots&\ldots&\ldots&\ldots\\ 0&\ldots&0&\lambda_k\ldots&0\\ \ldots&\ldots&\ldots&\ldots&\lambda_k\end{pmatrix}=\prod_{j=1}^k\lambda_j^{m_j}$$