<s>[INST] An exercise on the calculation of a function of operator The operator is given by
$$A=\begin{pmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 4
\end{pmatrix}$$
I have to write down the operator $$B=\tan(\frac{\pi} {4}A)$$
I calculate $$\mathcal{R} (z) =\frac{1}{z\mathbb{1}-A}=\begin{pmatrix} 
\frac{1}{z-1} & 0 & 0\\
\frac{1}{(z-1)^2} & \frac{1}{z-1} & 0\\
0 & 0 & \frac{1}{z-4}\end{pmatrix} $$
Now the B operator is given by:
$$B=\begin{pmatrix}
Res_{z=1}\frac{\tan(\frac{\pi}{4}z)}{z-1} & 0 & 0\\
Res_{z=1}\frac{\tan(\frac{\pi}{4}z)}{(z-1)^2} & Res_{z=1}\frac{\tan(\frac{\pi}{4}z)}{z-1} & 0\\
0 & 0 & Res_{z=4}\frac{\tan(\frac{\pi}{4}z)}{z-4}
\end{pmatrix} $$
For me the result should be
$$ B=\begin{pmatrix}
1 & 0 & 0\\
\frac{\pi}{2} & 1 & 0\\
0 & 0 & 0\end{pmatrix}$$
But the exercise gives as  solution:
$$ B=\begin{pmatrix}
1 & 0 & 0\\
\frac{\pi}{4} & 1 & 0\\
0 & 0 & 1\end{pmatrix}$$
Where is the error?
Thank you and sorry for bad English 
 [/INST] It would appear that there’s an error in either the problem statement or the solution.  
You can check your own answer by computing this via a consequence of the Cayley-Hamilton theorem: any analytic function of $A$ can be expressed as a quadratic polynomial in $A$, i.e., $\tan\left(\frac\pi4A\right) = aI+bA+cA^2$ for some unknown coefficients $a$, $b$, $c$. The eigenvalues of $A$ are obviously $1$, $1$ and $4$, so you can find these coefficients by solving the following system of linear equations $$a+b+c = \tan\frac\pi4 \\ b+2c = \frac\pi4\sec^2\frac\pi4 \\ a+4b+16 = \tan\pi$$ obtained by substituting the eigenvalues of $A$ into the above equation for $\tan\left(\frac\pi4A\right)$, as well as into its derivative in order to get a third independent equation in the case of the repeated eigenvalue. Solving these equations and computing the polynomial of $A$ produces $\pi/4$, not $\pi/2$ on the off-diagonal.
</s>