Determinants and diagonalizability of $n{\times}n$ matrix?

Question

Let $M$ be a $n {\times} n$- matrix such that $\det(M) \ne 0$ (not equal to 0). Which of the following statements is always true

A) $M$ is diagonalizable

B) $M$ has $n$ distinct eigenvalues

C) $0$ is an eigenvalue of M

D) All the eigenvalues are nonzero

E) $M$ has a set of eigenvectors that is a basis for $\Bbb R^n$

my answer would be D since determinants $\ne 0$ meaning we will have no eigenvalues of $0$

Answer 1

Since determinant of a matrix is same as product of the eigenvalues. So D is true whenever eigenvalue exist over the required field. For other cases, consider the $2 \times 2$ matrix $\begin{pmatrix}1 & 1\\ 0&1 \end{pmatrix}$. Here this matrix is not diagonalizable, so we eliminate A and D. Also $1$ is the eigenvalue with multiplicity two, so we eliminate B and C