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The operator $A$ is defined as $A(x,y) = (\frac{-3π}{4} x + \frac{π}{2} y, \frac{π}{2} x)$. I need to find the norm of the $cos(A)$ operator. I tried writing $cos(A)$ as a series in the hope that at some point $A^n$ would be zero, but this did not happen. Can you tell me what else I can try to apply?

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  • $\begingroup$ Since $A$ is a 2x2 matrix, you should be able to express $A^2$ as $aA+bI$ for some a and b. Maybe you can find a nice expression for $A^n$? $\endgroup$
    – JonathanZ
    Commented May 20 at 19:05
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    $\begingroup$ Diagonalise the matrix. $\endgroup$
    – copper.hat
    Commented May 20 at 19:13

1 Answer 1

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The powers (and hence the series) are invariant under similarity. You have $$ A=\begin{bmatrix} -3\pi/4&\pi/2\\ \pi/2&0\end{bmatrix} =\begin{bmatrix} -2&1/2\\1&1\end{bmatrix}\begin{bmatrix} -\pi&0\\0&\frac\pi4\end{bmatrix}\begin{bmatrix}-2/5&1/5\\2/5&4/5 \end{bmatrix}, $$ so $$ \cos A=\begin{bmatrix} -2&1/2\\1&1\end{bmatrix}\begin{bmatrix} -1&0\\0&\sqrt2/2\end{bmatrix}\begin{bmatrix}-2/5&1/5\\2/5&4/5 \end{bmatrix} =\frac1{10}\begin{bmatrix} -8+\sqrt2&4+2\sqrt2\\4+2\sqrt2&-2+4\sqrt2 \end{bmatrix}. $$

On second read, if you only need the norm, the above is not necessary for $A$ is selfadjoint. From this we know that the operator norm of $A$ is the largest eigenvalue in absolute value. The eigenvalues of $A$ are $-\pi$ and $\pi/4$, and so the eigenvalues of $\cos A$ are $-1$ and $\sqrt2/2$. Thus $\|\cos A\|=1$.

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    $\begingroup$ Good catch, thanks! I have edited the answer. $\endgroup$ Commented May 21 at 3:49

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