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I have $V$ which is the subspace of differentiable functions $\mathbb{R} \to \mathbb{R}$ spanned by set $S = \{\sin x, \cos x, x\sin x, x\cos x\}$.

I also have $T: V \to V$, $f \mapsto f'$ and I am supposed to find the rational canonical form of $T$.

So far, I know that $[T]_S = \begin{bmatrix}0 & -1 & 1 & 0\\1 & 0 & 0 & 0\\0 & 0 & 0 & -1\\0 & 0 & 1 & 0\end{bmatrix}$

Now, the characteristic polynomial for this matrix is $(x^2+1)^2$ and the minimal polynomial for this matrix is $x^2+1$.

The companion matrix corresponding to $x^2+1$ is $\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}$. Now, I originally thought that I simply make it into the following matrix:

$\begin{bmatrix}0 & -1 & 0 & 0\\1 & 0 & 0 & 0\\0 & 0 & 0 & -1\\0 & 0 & 1 & 0\end{bmatrix}$.

However, I am not entirely sure if this is the right format. The eigenvalues are $i$ and $-i$ and I am getting 1D eigenspace. Can someone help me from here? Thanks.

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    $\begingroup$ Do you know how to find the matrix of $T$ with respect to the given basis? You first need that matrix before you can put it in rational canonical form. $\endgroup$ – André 3000 Oct 30 '18 at 1:47
  • $\begingroup$ We know that the derivatives of each of those four elements are cos$x$, -sin$x$, sin$x + x$cos$x$, and cos$x - x$sin$x$, which should be represented into four columns, but I am not sure as to how to change that into matrix. $\endgroup$ – carlos mubuntu Oct 30 '18 at 1:49
  • $\begingroup$ Just write the coefficients as the columns. E.g., the first column is $\begin{pmatrix} 0\\ 1\\ 0\\ 0\end{pmatrix}$. If that's not clear, here are some links: 1, 2, 3. Or for the general procedure: 4. $\endgroup$ – André 3000 Oct 30 '18 at 1:54
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    $\begingroup$ That was relatively simple. Thanks. After I find the matrix, should I move on to find the rational canonical form by methods similar to this link? $\endgroup$ – carlos mubuntu Oct 30 '18 at 2:00
  • $\begingroup$ Yes, and in this case it suffices to just compute the minimal polynomial of the matrix $A$. So computing the characteristic polynomial is a good start. (In general you would have to compute all the invariant factors of $A$ in order to determine the rational canonical form.) If you figure it out, please post your answer below! $\endgroup$ – André 3000 Oct 30 '18 at 4:50

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