# The norm of the operator cos(A)

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?

• 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 $A^n$? Commented May 20 at 19:05
• Diagonalise the matrix. Commented May 20 at 19:13

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 $$\|\cos A\|=1$$.