Some matrices with real entries which are not diagonalizable over $\mathbb{R}$ are diagonalizable over the complex numbers $\mathbb{C}$

For instance,

$$A = \begin{pmatrix} 0 &-1 \\ 1 & 0 \end{pmatrix} $$

has $\lambda^2+1$ as characteristic polynomial. This polynomial doesn't factor over the reals, but over $\mathbb{C}$ it does. Its roots are $\lambda = \pm i$

Interpreting the matrix as a linear transformation $\mathbb{C}^2 \to \mathbb{C}^2$ it has eigenvalues $i$ and $-i$ and linearly independent eigenvectors $(1, -i)$ $(-i, 1)$ So we can diagonalize $A$

$$A = \begin{pmatrix} 0 &-1 \\ 1 & 0 \end{pmatrix}=\begin{pmatrix} 1 & -i \\ -i & 1 \end{pmatrix}\begin{pmatrix} i&0 \\0&-i \end{pmatrix}\begin {pmatrix} .5&.5i \\.5i&.5 \end{pmatrix}$$

But there exist real matrices which aren't diagonalizable even if complex eigenvectors and eigenvalues are allowed.

For example,

$$B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$

cannot be written as $U D U^{-1}$ with $D$ diagonal.

In fact, the characteristic polynomial is $\lambda^2$ and it has only one double root $\lambda=0$ However the eigenspace corresponding to the $0$ (kernel) eigenvalue has dimension 1.

$B \begin{pmatrix} v_1 \\v_2 \end{pmatrix} = \begin{pmatrix} 0 \\0 \end{pmatrix} \iff v_2 = 0 $ and thus the eigenspace is $ker(B) = span_{\mathbb{C}}\left \{ (1,0)^T \right\}$ with only one dimension.

There isn't a change of basis where $B$ is diagonal.