Prove $det(A)=\prod_{\lambda\in Spec(A)} \lambda^{m_\lambda}$ where $m_\lambda$ is the algebraical multiply, and $A\in Mat_3(\mathbb{C})$

Question

Prove $det(A)=\prod_{\lambda\in Spec(A)} \lambda^{m_\lambda}$ where $m_\lambda$ is the algebraical multiply, and $A\in Mat_3(\mathbb{C})$

I was thinking in use canonical form of jordan, but i cannot use this. (Formally i never seen the theory of this)

Then, for definition $det(A)=\sum \epsilon(\sigma)a_{1\sigma(1)}a_{2\sigma(2)}a_{3\sigma(3)}$

But i don't see how can i prove this by definition. Can someone help me?

Note: $Spec(A)$ is the set of eigenvalues of $A$

Answer 1

Hint: Write the eigen expansion, then take $det(M \Lambda M^{-1})$. Note: $det(M)=1/det(M^{-1})$