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 $e^{tA}$.
On
Let $u$ be the $4\times 1$ matrix with all entries equal to $1$. Let $u'$ be its transpose. Then $$ \begin{aligned} A &=uu'-I\ ,\\ A+I &=uu'\ ,\\ 4 &=u'u\ . \end{aligned} $$ Then $$ e^{tA}=e^{-tI+t(A+I)}=e^{-tI}\cdot e^{t(A+I)}=e^{-t}\cdot e^{t\,uu'}\ . $$ Now the powers of $uu'$ are easily computed. For instance $(uu')^2=uu'\ uu'=u\ u'u\ u'=4uu'$, inductively $(uu')^{n+1}=4^n\, uu'$. The exponential of $t\,uu'$ is then $$ \begin{aligned} \exp(tuu') &= I +\frac 1{1!}t\, uu' +\frac 1{2!}t^2\, 4\, uu' +\frac 1{3!}t^3\, 4^2\, uu' +\frac 1{4!}t^4\, 4^3\, uu' +\frac 1{5!}t^5\, 4^4\, uu' +\dots \\ &= I + \frac 14\left( \frac 1{1!}t\, 4 +\frac 1{2!}t^2\, 4^2 +\frac 1{3!}t^3\, 4^3 +\frac 1{4!}t^4\, 4^4 +\frac 1{5!}t^5\, 4^5 +\dots \right)\, uu' \\ &= I + \frac 14(e^{4t}-1)\,uu' \ . \end{aligned} $$ The above $uu'$ is the matrix with ones as entries.
From here we get the needed formula, as confirmed by sage:
sage: A = matrix( QQ, 4,4, [0,1,1,1, 1,0,1,1, 1,1,0,1, 1,1,1,0] )
sage: A
[0 1 1 1]
[1 0 1 1]
[1 1 0 1]
[1 1 1 0]
sage: var('t');
sage: exp(t*A)
[1/4*(e^(4*t) + 3)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t)]
[1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) + 3)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t)]
[1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) + 3)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t)]
[1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) + 3)*e^(-t)]
sage: latex(_)
$$ \left(\begin{array}{rrrr} \frac{1}{4} \, {\left(e^{\left(4 \, t\right)} + 3\right)} e^{\left(-t\right)} & \frac{1}{4} \, {\left(e^{\left(4 \, t\right)} - 1\right)} e^{\left(-t\right)} & \frac{1}{4} \, {\left(e^{\left(4 \, t\right)} - 1\right)} e^{\left(-t\right)} & \frac{1}{4} \, {\left(e^{\left(4 \, t\right)} - 1\right)} e^{\left(-t\right)} \\ \frac{1}{4} \, {\left(e^{\left(4 \, t\right)} - 1\right)} e^{\left(-t\right)} & \frac{1}{4} \, {\left(e^{\left(4 \, t\right)} + 3\right)} e^{\left(-t\right)} & \frac{1}{4} \, {\left(e^{\left(4 \, t\right)} - 1\right)} e^{\left(-t\right)} & \frac{1}{4} \, {\left(e^{\left(4 \, t\right)} - 1\right)} e^{\left(-t\right)} \\ \frac{1}{4} \, {\left(e^{\left(4 \, t\right)} - 1\right)} e^{\left(-t\right)} & \frac{1}{4} \, {\left(e^{\left(4 \, t\right)} - 1\right)} e^{\left(-t\right)} & \frac{1}{4} \, {\left(e^{\left(4 \, t\right)} + 3\right)} e^{\left(-t\right)} & \frac{1}{4} \, {\left(e^{\left(4 \, t\right)} - 1\right)} e^{\left(-t\right)} \\ \frac{1}{4} \, {\left(e^{\left(4 \, t\right)} - 1\right)} e^{\left(-t\right)} & \frac{1}{4} \, {\left(e^{\left(4 \, t\right)} - 1\right)} e^{\left(-t\right)} & \frac{1}{4} \, {\left(e^{\left(4 \, t\right)} - 1\right)} e^{\left(-t\right)} & \frac{1}{4} \, {\left(e^{\left(4 \, t\right)} + 3\right)} e^{\left(-t\right)} \end{array}\right) $$
On
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 $x=A$) $$ e^{tA}=\frac{e^{3t}-e^{-t}}{4}(A+I)+e^{-t}I. $$
On
$A^2=3I+2A$, 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....
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*}