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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?

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See this math.stackexchange.com/questions/211648/… –  PAD Nov 1 '12 at 7:43
    
Thank you. I tried searching for a little while to see if someone else had already asked the question but didn't manage to to find that. –  chris Nov 1 '12 at 7:46
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