jordan canonical form with direct product?

I met some problems when solving Jordan canonical forms. Here are two problems:

1. Let $f: K^3\to K^3$ be a map in JCF having the matrix: $$\begin{pmatrix} -1 & 1 & 0\\ 0 &-1&1\\ 0&0&-1\\ \end{pmatrix}$$ Find the JCF of the map $\bigwedge^2 f$.

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

Well, I know how to find the JCF for a matrix(at least 3 by 3, for example the first matrix is already in the Jordan blocks form). However, I don't know what does the $\bigwedge^2 f$ and $f\otimes f$ mean.

1. Could any body help me to explain what is and how the "$\bigwedge^2 f$ and $f\otimes f$" changes the matrix?

2. If the result matrix is too complicated, is there any quick way to find the required JCF? I mean, if the result matrix is 9 by 9(I guess), is there any other way to find the JCF other than find the eigenvalues of the 9 by 9 matrix, and follow the routine procedure?

Thank you very much!

• my naive viewpoint is that $f \otimes f$ is simply formed by sticking $f$ in each component of $f$. It is a $9 \times 9$ matrix. The $\Lambda^2 f$ is likely an antisymmetric object based on $f$, but, I'm not certain, do you have a reference? – James S. Cook Jan 6 '15 at 2:33

The first map is just the induced map on the $2$-graded component of the exterior algebra on $K^3$, namely $\Lambda^2(K^3)$. The second is the map on the $2$ tensors in the tensor algebra, i.e. elements of $K^3\otimes K^3$.

On the exterior product we have the diagonal action, so determine the action on the basis in which your matrix is written. If the matrix basis is $\{e_1,e_2,e_3\}$ then the basis for $\Lambda^2(f)$ is naturally $\{e_1\wedge e_2, e_2\wedge e_3, e_3\wedge e_1\}$, and write down the matrix relative to this. It should be $3\times 3$ since $\dim \Lambda^2(K^3)=3$.

In the first example, note that

$$\Lambda^2(f)(e_1\wedge e_2)=f(e_1)\wedge f(e_2)$$

computing, we see from the JCF this is

$$(-e_1)\wedge (e_1-e_2)=e_1\wedge e_2$$

similarly for the other two basis vectors we get

$$\begin{cases}\Lambda^2(f)(e_2\wedge e_3)=(e_1-e_2)\wedge (e_2-e_3)=e_1\wedge e_2+e_2\wedge e_3+e_3\wedge e_1 \\ \Lambda^2(f)(e_3\wedge e_1)=(e_2-e_3)\wedge (-e_1)= e_1\wedge e_2+e_3\wedge e_1 \end{cases}$$

so the matrix relative to this basis is

$$\Lambda^2(f) = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}$$

and you can compute the JCF from this.

On the tensor product, you'll get quite a large matrix--$9\times 9$ as you noted--but it's mercifully sparse thanks to all the zeroes. You can easily compute the action of $f\otimes f$ on the basis $\{e_i\otimes e_j\}_{i,j=1}^3$ and then right down the linear transformation relative to this.

As an example, note that

$$(f\otimes f) (e_1\otimes e_1)=f(e_1)\otimes f(e_1)=e_1\otimes e_1.$$

This indicates the first column of $f\otimes f$ relative to this basis is

$$\begin{pmatrix}1 \\ 0 \\ 0 \\0\\0\\0\\0\\0\\0\end{pmatrix}$$

Edit: Why not, I'll do more computations.

The others are also straightforward:

$$\begin{cases} (f\otimes f)(e_1\otimes e_2)=(-e_1)\otimes (e_1-e_2)=-(e_1\otimes e_2)+(e_1\otimes e_2)\\ (f\otimes f)(e_1\otimes e_3)=(-e_1)\otimes (e_2-e_3)=-(e_1\otimes e_2)+(e_1\otimes e_3)\\ (f\otimes f)(e_2\otimes e_2)=(e_1-e_2)\otimes (e_1-e_2)=(e_1\otimes e_1)-(e_1\otimes e_2)-(e_2\otimes e_1)+(e_2\otimes e_2)\\ (f\otimes f)(e_3\otimes e_3)=(e_2-e_3)\otimes(e_2-e_3)=(e_2\otimes e_2)-(e_2\otimes e_3)-(e_3\otimes e_2)+(e_3\otimes e_3)\\ (f\otimes f)(e_3\otimes e_2)=(e_2-e_3)\otimes(e_1-e_2)=(e_2\otimes e_1)-(e_2\otimes e_2)-(e_3\otimes e_1)+(e_3\otimes e_2)\\ (f\otimes f)(e_3\otimes e_1)=(e_2-e_3)\otimes(-e_1)=(e_3\otimes e_1)-(e_2\otimes e_1)\\ (f\otimes f)(e_2\otimes e_1)=(e_1-e_2)\otimes (-e_1)=(e_2\otimes e_1)-(e_1\otimes e_1)\\ (f\otimes f)(e_2\otimes e_3)=(e_1-e_2)\otimes (e_2-e_3)= (e_1\otimes e_2)-(e_2\otimes e_2)+(e_2\otimes e_3)-(e_1\otimes e_3) \end{cases}$$

giving a whopping matrix of

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

As for speed, I don't know of a much faster way other than noting the sparseness makes doing the usual procedure much faster by starting with a JCF tensor rather than doing a generic transformation, tensoring, then trying JCF.

Note: You are not obligated to order your basis in the same way, I did that in my head, so I just took the ordering that came out when I was doing them. Certainly reordering will make it easier to deal with. Of course, you can get the eigenvalues most easily from the reduced echelon form using only row permutation and non-scaling combinations, since upper-triangular form is enough for the eigenvalues.

• Nice answer! I'm curious, do you happen to know a source where the connection between the exterior product of matrices (or linear transformations) and the cofactor expansion is explicitly given. – James S. Cook Jan 6 '15 at 3:33
• @JamesS.Cook If I'm remembering right, you can easily derive cofactors from the definitions of the determinant of an exterior $k$-form and the exterior product. Writing the matrix $$M=\begin{pmatrix} | & \ldots & | \\ v_1 & \ldots & v_k \\ | & \ldots & | \end{pmatrix}$$ Then writing $$v_i=\sum_{j=1}^k v_{ij}e_j$$ you can derive the cofactor formula, since $\det M$ is just the norm of the $k$-form $(v_1\wedge\ldots \wedge v_k)\in \Lambda^k(K^n)$. In the classical case, of course, $k=n$, but this isn't strictly necessary. – Adam Hughes Jan 6 '15 at 3:36