Basis of the matrices with only non diagonalizable matrices

Question

Is it possible to find a basis of $M_n(\mathbb{R})$ that only has non diagonalisable matrices ?

I'm looking for a rather easy example, or a proof of the (non-)existence.

Answer 1

This is always possible: let $E_{ij}$ be the matrix with $1$ in the $i, j$ spot and $0$ elsewhere (notice $\{E_{ij} \}$ is the standard basis of $M_n(\mathbb{R})$). For $i \ne j$, $E_{ij}^2 = 0$, hence is nilpotent and not diagonalizable, so $N = \{E_{ij} \mid i \ne j\}$ is a linearly independent set of $n^2 - n$ non-diagonalizable matrices.

Next, note that for any $i$, it is always possible to choose $j < k$ such that $E_{ii} + E_{jk}$ has rank $2$, as long as $n \ge 3$. Since $E_{ii} + E_{jk}$ is triangular with characteristic polynomial $\lambda^{n-1}(\lambda - 1)$, the algebraic multiplicity of $0$ is $n - 1$, but the geometric multiplicity is $\dim \ker (E_{ii} + E_{jk}) = n - 2$. Thus $E_{ii} + E_{jk}$ is not diagonalizable, and $N \cup \{E_{ii} + E_{j_ik_i} \mid 1 \le i \le n\}$ is a set of $n^2$ matrices with the same span as $\{E_{ij}\}$, hence is a basis of $M_n(\mathbb{R})$.

Explicitly, one can take as basis $N \cup \{E_{22} + E_{13}\} \cup \{E_{33} + E_{12}\} \cup \{E_{ii} + E_{23} \mid i \ne 2, 3\}$.

For $n = 2$, one can check that

$$\begin{bmatrix}1 & 1 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix}0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix}-1 & 4 \\ -1 & 3 \end{bmatrix}$$

is a basis of $M_2(\mathbb{R})$, none of which are diagonalizable (the last is similar to the first).