So if I have an $n\times n$ matrix $A$ that is diagonalizable then how do I determine $\mbox{tr}(A)$ and $\det(A)$ as functions of the eigenvalues of the matrix?
For $\det(A)$, I know the formula for eigenvalues is $\det(A−tI)$ which would be $$(-1)^n(t-\lambda_1)\dots (t-\lambda_n)$$ where $\lambda_i$ are the eigenvalues of $A$.
For $\mbox{tr}(A)$ could I also do something similar to that? I'm just not sure how this question works.