# Diagonalizing Matrices of Operators/Finding Jordan Normal form

Suppose we have bases on an inner product space $V$: $\beta = \{ v_1, v_2 \}$ and $\beta' = \{v_1, v_1 + v_2 \}$ such that $\|v_1\| = 1$, $\|v_1 + v_2\|=1$, and $\langle v_1, v_1 + v_2 \rangle = 0$. Suppose that $T: V \to V$ is a linear operator and that the matrix of $T$ in $\beta$ is:

$$[T]_{\beta} = \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}.$$

Then, the matrix of $T$ in $\beta'$ is:

$$[T]_{\beta'} = \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}.$$

But these matrices are not transformable into one another by a linear change of coordinates! So do the theorems about Jordan Normal Forms/Diagonalizabilty not hold for matrices of linear operators/tensors?

The matrices are similar and you have just demonstrated that. Note that both matrices have the same (distinct!) eigenvalues and any two matrices which have the same eigenvalues, all distinct, are in fact similar. This does not contradict the Jordan normal form because a $2 \times 2$ Jordan block is of the form
$$\begin{pmatrix} a & 1 \\ 0 & a \end{pmatrix}$$
$$\begin{pmatrix} a & 1 \\ 0 & b \end{pmatrix}.$$