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I came across this question, asked in a competitive exam. It is as follows.

Given a matrix M = \begin{bmatrix}2&1\\1&2\end{bmatrix} what is the value of cos(πM/6)?

I've tried series expansion but I think there is an alternative way doing it, any help is appreciated.

Options given are

\begin{bmatrix}1/2&1\\1&1/2\end{bmatrix} \begin{bmatrix}\sqrt3/4&-\sqrt3/4\\-\sqrt3/4&\sqrt3/4\end{bmatrix} \begin{bmatrix}\sqrt3/4&\sqrt3/4\\\sqrt3/4&\sqrt3/4\end{bmatrix} \begin{bmatrix}1/2&\sqrt3/2\\\sqrt3/2&1/2\end{bmatrix}

user1844
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    did you try diagonalizing $M$? –  Feb 16 '18 at 06:13
  • But still, it would contain a pi factor right, but the options does not have them at all, may be I am making a mistake, please help me out. – user1844 Feb 16 '18 at 06:43
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    No one can tell if you make a mistake if you do not show us what you did..... –  Feb 16 '18 at 07:01
  • http://infohost.nmt.edu/~iavramid/notes/mp2-5.pdf page 58 – Trevor Gunn Feb 16 '18 at 07:08
  • Observe that $\cos M=A^{-1}\cos(AMA^{-1})A$. So we can reduce to a diagonal form. – Angina Seng Feb 16 '18 at 07:13
  • @ John Ma diagonalization should leave the off diagonal entries zero, even after doing the power series expansion of cosine, right. Multiplying with \pi /6 is going to introduce a \pi ^2 term in the power series expansion, but none of the options contains a '\pi' factor in them, this was my doubt. – user1844 Feb 16 '18 at 07:16
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    Your intuition does not make sense since the series is infinite. For example we have $$0=\sin \pi = \pi - \frac{\pi^3}{3!} + \frac{\pi^5}{5!} - \frac{\pi^7}{7!} + \cdots $$ –  Feb 16 '18 at 07:20

1 Answers1

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It is not clear from the question whether you are looking for the series expansion solution or an alternative.

For the series expansion, remember that

$$\cos(M) = I - \frac{M^2}{2!} + \frac{M^4}{4!} - ...$$

and that $M^n$ can be found by diagonalising.

An alternative solution has been given in part by user akhmeteli on the Physics SE.

It is easy to show that $3$ is an eigenvalue of $M$ with eigenvector $\left( \begin{matrix} 1 \\ 1 \end{matrix} \right)$. By virtue of the series expansion, we know that this is an eigenvector of $\cos(\frac{\pi M}{6})$ and so the corresponding eigenvalue is $\cos(\frac{\pi 3}{6}) = 0$. Thus we are looking for a matrix with determinant $0$ and with an eigenvalue (corresponding to our eigenvector) of $0$. Only one of the solutions satisfies this condition.

mauiaw1
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