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Given the matrix

$$A = \begin{pmatrix} \cos x & \sin x\\ -\sin x & \ \cos x \end{pmatrix}$$

I have to express it in rational form

The characteristic polynomial is $y^2 - 2y\cos x+1 $

Accordingly, the rational form $R$ if $x$ is neither $0$ or $2 \pi$ is

$$ R = \begin{pmatrix} 0 & -1\\ 1 & \ 2 \cos x\end{pmatrix}$$

Please suggest if it is correct.

If $x = 0$ or $2 \pi$, then Min poly of $A = x-1$.

Then, it seems the rational form is

$$ R = [1]$$ Please help.

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On the internet, using all caps is equivalent to shouting, which is very rude. Please don't do it again. – Zev Chonoles Apr 19 '12 at 21:01
@ZevChonoles: Thanks for the edit ! – preeti Apr 19 '12 at 21:14
What do you mean by 'the rational form of a matrix'? – copper.hat Apr 20 '12 at 3:05
rational canonical form – Graphth Apr 20 '12 at 20:05
Your rational canonical form is incorrect if $x=\pi$; in that case, the characteristic polynomial is $y^2 +2y + 1$, which factors; the minimal polynomial is $y+1$. – Arturo Magidin Apr 20 '12 at 20:09
up vote 2 down vote accepted

The characteristic polynomial of $$\left(\begin{array}{rr} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{array}\right)$$ is, as you note, $t^2 -2(\cos\theta)t + 1$.

The discriminant of this polynomial is $$4(\cos^2\theta) - 4 = 4(\cos^2\theta - 1).$$ Therefore, the discriminant is always nonpositive, and is equal to zero if and only if $\cos^2\theta = 1$, if and only if $\theta = n\pi$ for some $n\in\mathbb{Z}$.

In particular, if $\theta\neq n\pi$ then the characteristic polynomial is irreducible over $\mathbb{Q}$, so the minimal polynomial agrees with the characteristic polynomial, and so the rational canonical form is just the companion matrix of the characteristic polynomial (just as you wrote): $$\left(\begin{array}{rr} 0 & -1\\ 1 & 2\cos\theta \end{array}\right).$$

If $\theta=n\pi$, then $A$ is one of the following matrices: $$\begin{align*} A &= \left(\begin{array}{rr} 1 & 0\\ 0 & 1 \end{array}\right) &\text{if }n\text{ is even,}\\ A &=\left(\begin{array}{rr} -1 & 0\\ 0 & -1 \end{array}\right) &\text{if }n\text{ is odd.} \end{align*}$$

Both of these matrices are already in rational canonical form.

(Your error: the minimal polynomial is indeed $t-\cos\theta$ in this situation; since the characteristic polynomial is then the square of the minimal polynomial, the rational canonical form will have two blocks, each associated to a degree 1 polynomial; not just a single block. Remember that the exponent of the irreducible factor on the minimal polynomial gives you the size of the largest block, in this case $1\times 1$; but the sum of the sizes of the blocks has to add up to the exponent of the irreducible factor in the characteristic polynomial, in this case $2$. So you need two $1\times 1$ blocks here. )

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Thanks a lot for the help. – preeti Apr 22 '12 at 21:43

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