Finding a basis where the rational canonical form holds

Question

The matrix I have to consider is $A= \begin{bmatrix}x-\frac{1}{2}&\frac{1}{2}&\frac{1}{2}&-\frac{1}{2}\\-\frac{1}{2}&x-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}\\\frac{1}{2}&\frac{1}{2}&x-\frac{1}{2}&-\frac{1}{2}\\-\frac{1}{2}&\frac{3}{2}&-\frac{1}{2}&x-\frac{1}{2}\end{bmatrix}$.
I know the Smith normal form of $xI-A$ is $\begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&x^4-2x^3+2x^2-1\end{bmatrix}$, and the rational form of $A$ is $\begin{bmatrix} 0&0&0&1 \\ 1&0&0&0 \\ 0&1&0&-2\\0&0&1&2\end{bmatrix}$. But how can I obtain the a basis in $\mathbb{Q}^4$ where the rational form holds? I tried a matlab function [V,J]=jordan(A), but it doesn't work(I don't know why). Can you explain how to get such basis?

Answer 1

I’ll start with the general picture for an endomorphism $\alpha : V \to V$ where $V$ is a finite-dimensional $k$-vector space $V$. Recall that we define a $k[x]$-module structure on $V$ by letting $xv = \alpha(v)$ and otherwise continuing the $k$-multiplication.

We now want to find a finite presentation of $V$ as a $k[x]$-module. To this end, we choose a basis of $V$ as a $k$-vector space, say $\mathcal B = \{ v_1, \dots, v_n \}$ and denote by $A$ the matrix representing $\alpha$ in this basis. (If you only want to think about matrices, simply take $\mathcal B$ to be the standard basis of $k^n$, in which case $\alpha = A$ from now on.) We can let $A$ act on elements of $k[x]^n$ as well (by usual matrix multiplication as we can multiply polynomials by elements of $k$) and we can then write down the exact sequence $$ k[x]^n \xrightarrow{xI - A} k[x]^n \xrightarrow{~~~\varphi~~~} V \longrightarrow 0 $$ where $\varphi$ is the map with $e_i \mapsto v_i$, or more explicitly $$ (p_1, \dots, p_n)^T \mapsto (p_1(\alpha) v_1 + \cdots + p_n(\alpha) v_n). $$ (If you’re not familiar with exact sequences, this means that $\varphi$ is surjective and the image of $xI - A$ is the kernel of $\varphi$. Put differently, $V \cong k[x]^n/(\operatorname{im} S)$ where the isomorphism is induced from $\varphi$. Note that this implies that the matrix $xI - A$ determines $V$ as a $k[x]$-module up to isomorphism; this is why finite presentations are useful: they turn an abstract object (like the module $V$) into a concrete matrix.) I omit the proof of exactness.

The algorithm computing the Smith normal form $S$ gives two additional invertible matrices $P$ and $Q$ (with polynomial entries) such that $S = P(xI - A)Q$. Replacing $xI - A$ by $P^{-1}SQ^{-1}$ in the sequence above and changing the bases in the two copies of $k[x]^n$, we get the new finite presentation $$ k[x]^n \xrightarrow{~~~S~~~} k[x]^n \xrightarrow{~~\varphi \circ P^{-1}~~} V \longrightarrow 0. $$ (The exact position of $P^{-1}$ is much less surprising if you remember that we read function composition from right to left but the arrows go from left to right. In reality, $P^{-1}$ was and still is “closer” to the second copy of $k[x]^n$. As for $Q^{-1}$, it “vanished” into the basis for the first copy of $k[x]^n$ and we don’t need to care about it anymore.)

The images of the standard basis vectors of $k[x]^n$ under $\varphi \circ P^{-1}$ still are a generating set of $V$ as a $k[x]$-module, but in general, they won’t be a $k$-basis anymore. (In fact, the standard basis vectors corresponding to a $1$ on the diagonal of $S$ will go to zero.) Instead, you need to take only the standard basis vectors of $k[x]^n$ corresponding to the non-$1$ diagonal entries of $S$. The image of each of them (under $\varphi \circ P^{-1}$) generates one of the $x$-invariant subspaces of $V$ we’re interested in. You get a basis of this subspace by applying $x$ (equivalently, $\alpha$) to this vector.

In your case, a possible choice for $P^{-1}$ with $P^{-1}SQ^{-1} = xI - A$ is $$ P^{-1} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & x-1 & 1 \\ x^2 - 2x + 1 & -x^2 & -x^2 + x - 1 & -x \end{pmatrix} $$ (keep in mind that a computer algebra system will likely return $P$, i.e. the inverse of this) and $S$ is as in your question, i.e. $$ S = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 &x^4 - 2x^3 + 2x^2 - 1 \end{pmatrix}. $$

Applying $\varphi \circ P^{-1}$ to the last basis vector (which is the only one where the Smith normal form has a non-1 diagonal entry) gives $$ u_1 = (\varphi \circ P^{-1})(e_4) = \varphi(e_3-x e_4) = e_3 - Ae_4 = (-1/2, -1/2, 1/2, -1/2)^T. $$ (On the left, the $e_i$ are the standard generators of $k[x]^4$; on the right, $e_i$ are the basis vectors in $k^4$ because that is the vector space and basis with which we started in your particular example.) Applying $A$ to this, gives \begin{align*} u_2 &= Au_1 = (-1/2, -1/2, 1/2, 1/2)^T \\ u_3 &= A u_2 = (0, 0, 1, 1)^T \\ u_4 &= A u_3 = (0, 1, 1, 1)^T \end{align*} Note that applying $A$ to $u_4$ gives $$ A u_4 = (-1/2, 3/2, 1/2, -1/2)^T = u_1 - 2u_3 + 2u_4. $$ Together with the equations defining $u_2, u_3, u_4$ this implies that your rational form is indeed the representing matrix of $A$ in this new basis.