# Find Jordan canonical form with Kronecker product

Let $f: K^3\to K^3$ be a map in Jordan canonical form having a matrix $$\begin{pmatrix} 1 & 1 & 0\\ 1 &0&1\\ 0&1&1\\ \end{pmatrix}$$ Find the JCF of the map $f\otimes f$.

My question is, do I really have to do the JCF for a $9 \times 9$ matrix? or I need to use some property of the relationship between JCF of $f\otimes f$ and the JCF of $f$?

• What do you mean by "a map in Jordan canonical form"? Certainly that matrix is not a Jordan canonical form. – Marc van Leeuwen Apr 3 '15 at 6:08

That matrix is real symmetric, therefore diagonalisable, as it happens with distinct integral eigenvalues, forming the set $\{-1,1,2\}$. Since the Kronecker product of diagonal matrices is diagonal, it follows that $f\otimes f$ is also diagonalisable, with multiset of eigenvalues $$\{\!\{\,ab\mid \, a,b\in \{\!\{-1,1,2\}\!\}\,\}\!\} = \{\!\{-2,-2,-1,-1,1,1,2,2,4\}\!\}.$$