Let $f\colon K^3\to K^3$ be a map in Jordan canonical form having matrix $ f=\begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0& -1 \end{bmatrix} $ What is $f\otimes f$? What is $f\wedge f$?
I'm just trying to make sure that I'm understanding exactly what to do with each part of this problem. First, the dimension of $f\otimes f$ is $9$ because $f$ is a $3$-dimensional tensor. I think this means that I will end up with a block matrix where each block is the appropriate multiple of $f$: $ f\otimes f = \begin{bmatrix} -f & f & 0 \\ 0 & -f & 0 \\ 0 & 0 & -f \end{bmatrix} $ Is this understanding correct?
For $f\wedge f$, I'm not as sure what to do. The dimension of $f\wedge f$ is $\binom{3}{2}=3$, so I think I'm supposed to multiply $f$ by some multiple of itself. What exactly am I supposed to do?