Easy ways to calculate matrix exponential of a particular $4\times 4$ matrix Wondering how to find the matrix exponential of the following matrix without having to do through the long process of finding eigenvalues/eigenvectors and Jordan forms.  Is there a quicker way to do it using Sine/Cosine ?
If 
$$A= \begin{pmatrix}0 & 1 & 1 & 1 \\ 
                     1 & 0 & 1 & 1 \\
                     1 & 1 & 0 & 1 \\
                     1 & 1 & 1 & 0 \\
\end{pmatrix}$$ 
Find $e^{tA}$.
 A: Let $\boldsymbol A = \boldsymbol U - \boldsymbol I_4$. Simple calculations show that $\boldsymbol U^n = 4^{n-1} \boldsymbol U$. Now for every $n \in \mathbb N^*$:
\begin{align*}
\boldsymbol A^n =( \boldsymbol U - \boldsymbol I_4)^n &= \sum_0^n \binom n j \boldsymbol U^{j} (-1)^{n-j} \boldsymbol I \\
&=(-1)^n \boldsymbol I + \sum_1^n \binom n j 4^{j-1}(-1)^{n-j} \boldsymbol U \\
&= (-1)^n\boldsymbol I + \frac 14 \sum_0^n \binom n j 4^j (-1)^{n-j} \boldsymbol U - (-1)^n \boldsymbol U /4 \\
&= \frac 14 (4-1)^n \boldsymbol U + (-1)^n\boldsymbol I - (-1)^n \boldsymbol U/4 \\
&= \frac 14 \cdot (3^n - (-1)^n) \boldsymbol U + (-1)^n\boldsymbol I  
\end{align*}
Therefore,
\begin{align*}
\exp(t\boldsymbol A) &= \sum_0^\infty \frac 1 {n!} t^n\boldsymbol A^n \\
&= \sum_0^\infty \frac {t^n}{4n!} (3^n \boldsymbol U - (-1)^n \boldsymbol U + 4(-1)^n\boldsymbol I) \\
&= \frac 14 (\mathrm e^{3t} - \mathrm e^{-t}) \boldsymbol U + \mathrm e^{-t}\boldsymbol I.
\end{align*}
A: Another way to do it is to use the minimal polynomial and Lagrange interpolation. Calculating $A^2$ we notice that $A^2=3I+2A$, hence, $A^2-2A-3I=0$. It means that
$$
\pi_A(x)=x^2-2x-3=(x+1)(x-3)
$$
annihilates $A$, i.e. $\pi_A(A)=0$. Let's decompose the function $e^{tx}$ as
$$
e^{tx}=q(x)\pi_A(x)+r(x).
$$
If we manage to do that (with $q$, $r$ analytical near zeros of $\pi_A$) then
$$
e^{tA}=q(A)\pi_A(A)+r(A)=q(A)\cdot 0+r(A)=r(A).
$$
Since $\pi_A$ is of degree 2 it is enough to search for $r$ of degree 1 in the form $r(x)=ax+b$. To find $a,b$ we can use interpolation conditions at $x=-1$ and $x=3$:
$$
e^{-t}=r(-1)=-a+b,\qquad e^{3t}=r(3)=3a+b.
$$
Solving the system of 2 equations and 2 unknowns gives
$$
a=\frac{e^{3t}-e^{-t}}{4},\qquad b=\frac{e^{3t}-e^{-t}}{4}+e^{-t},
$$
that is 
$$
r(x)=ax+b=\frac{e^{3t}-e^{-t}}{4}(x+1)+e^{-t}.
$$
Finally (with $x=A$)
$$
e^{tA}=\frac{e^{3t}-e^{-t}}{4}(A+I)+e^{-t}I.
$$
A: $A^2=3I+2A$, so $e^{tA}$ is a linear combination of $I$ and $A$:
$$e^{tA}=f(t)I+g(t)A.$$
But the derivative of $e^{tA}$ is $Ae^{tA}$:
$$f'(t)I+g'(t)A=f(t)A+g(t)(2A+3I)=3g(t)I+(f(t)+2g(t))A.$$
We get the system of differential equations
\begin{align}
f'(t)&=3g(t)\\
g'(t)&=f(t)+2g(t)
\end{align}
with initial conditions $f(0)=1$ and $g(0)=0$. We get the second-order
equation for $g(t)$:
$$g''(t)-2g'(t)-3g(t)=0.$$
Solve this, use initial conditions to find arbitrary constants, and extract
$f(t)=g'(t)-2g(t)$ from it....
