Trace, determinant and eigenvalues for non-diagonalizable matrix Is the trace and determinant equal to the sum and product, respectively, of the eigenvalues even if a matrix is not diagonalizable? 
The proof I've seen for the trace equaling the sum of the eigenvalues and the determinant equaling the eigenvalues assumes a matrix can be written as $PDP^{-1}$. But if it is not diagonalizable, then we can't possibly write $PDP^{-1}$. So what happens then?
 A: This answer assumes the matrices are taken over $\mathbb C$.
Yes, the statement is still true even if the matrix isn't diagonalizable.
For the proof you saw it is sufficient that $D$ can be taken to be an upper triangular matrix (and it can be taken in such a way, this is Schur's Decomposition Theorem). This is enough because its diagonal entries will be the eigenvalues of the starting matrix.
Jordan Canonical Form is also sufficient, but Schur's Decomposition is a weaker condition.

For completeness I'll add the proofs here. 
Let $n\in \mathbb N$ and $A\in \mathcal M_n(\mathbb C)$. Let $\lambda _1, \ldots ,\lambda _n$ be the eigenvalues of $A$. The characteristic polynomial $p_A(z)$ of $A$ is $\color{grey}{p_A(z)=}(z-\lambda _1)\ldots (z-\lambda _n)$.
Schur's Decomposition guarantees the existence of an invertible matrix $P$ and an upper triangular matrix $U$ such that $A=PUP^{-1}$ and $U$'s diagonal entries are exactly $\lambda _1, \ldots ,\lambda _n$.
Since similarity preserves the characteristic polynomial, it follows that the characteristic polynomial $p_U(z)$ of $U$ is $\color{grey}{p_U(z)=}(z-\lambda _1)\ldots (z-\lambda _n)$, therefore $U$ and $A$ have the same eigenvalues with the same algebraic multiplicity.
From the fact that $U$'s diagonal entries are $\lambda _1, \ldots ,\lambda _n$ it follows that the trace of $U$ is the sum of the eigenvalues of $A$ and the determinant of $U$ is the product of the eigenvalues of $A$. 
Trace properties yield the following $$\text{tr}(A)=\text{tr}\left(PUP^{-1}\right)=\text{tr}\left(UP^{-1}P\right)=\text{tr}(U),$$ thus proving that the sum of the eigenvalues of $A$ equals $\text{tr}(A)$.
Similarly for the determinant it holds that $$\det(A)=\det\left(PUP^{-1}\right)=\det\left(P\right)\det\left(U\right)\det\left(P^{-1}\right)=\det(U),$$
hence the product of teh eigenvalues of $A$ equals the determinant of $A$.
A: Yes. If the matrix is not diagonalizable then still it can be written in Jordan canonical form. And trace relation and determinant relation can be determined using this.
