find the trace and the determinant of an nxn diagonizable matrix

Question

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Question: if A is a nxn diagonizable matrix, determine tr(A) and Det(A) as functions of eigenvalues of of the matrix?

Answer 1

If we write the Jordan Normal Form for the matrix $A$ we get:

$$A=P^{-1}JP$$

and $J$ has all eigenvalues on the diagonal but $J$ not necessearily is a diagonal matrix.

So $\det(A)=\det(J)$ and $\det(J)$ is the product of the eigenvalues.

Now using the known relation $\mbox{tr} (AB)=\mbox{tr} (BA)$ we have:

$$\mbox{tr} (A)=\mbox{tr} (P^{-1}JP)=\mbox{tr} (JPP^{-1})=\mbox{tr} (J)$$

and $\mbox{tr} (J)$ is the sum of eigenvalues.