Diagonalize the matrix $A$ and compute $e^{At}$

Question

I have the matrix

$A = \pmatrix{ -21 & 36 & -4\\ -14 & 23 & -2 \\ -8 & 8 & 3}$

And I am asked to diagonlaize it. I have find that the eigen values are

$\lambda_1 = 3 \\ \lambda_2 = 1+2i \\ \lambda_3 = 1-2i$

By calculating the corresponding eigenvectors I mind that my matrix $M$ is

$M = \pmatrix{ 1 & 0.5 & 1.5 \\ 1 & 0.5 & 1 \\ 3 & 1 & 1}$

Where $M = (\vec e_1| \vec a + \vec b | \vec a - \vec b)$, $a$ and $b$ are the real and imaginary part of my eigenvectors.

I can then calculate $\tilde{A} = M^{-1}AM$ which gives me

$\tilde{A} = \pmatrix{3 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 2 & 1}$

Now using this matrix I need to calculate the exponential map $e^{\tilde{A}t}$

which is given as

$\pmatrix{ e^{3t} & 0 & 0 \\ 0 & e^tcos(2t) & -e^tsin(2t) \\ 0 & e^tsin(2t) & e^tcos(2t)}$

but I don't know how to get that. Like I can't see how you go from our matrix $\tilde{A}$ to that matrix for $e^{\tilde{A}t}$

I just need someone to explain how to get the matrix for $e^{\tilde{A}t}$

Answer 1

Let$$A_1=\begin{pmatrix}3&0&0\\0&1&0\\0&0&1\end{pmatrix}\text{ and }A_2=\begin{pmatrix}0&0&0\\0&0&-2\\0&2&0\end{pmatrix}.$$Then $\tilde{A}=A_1+A_2$ and therefore, $t\tilde A=tA_1+tA_2$. Now, in general it is false that$$\exp(M+N)=\exp(M)\exp(N),$$but it is true when $M$ and $N$ commute. So, since $A_1$ and $A_2$ do commute$$\exp\left(t\tilde A\right)=\exp(tA_1)\exp(tA_2)$$and these exponentials are easy to compute. Note that, for each $n\in\mathbb N$,$$(tA_1)^n=\begin{pmatrix}(3t)^n&0&0\\0&t^n&0\\0&0&t^n\end{pmatrix}$$and if you compute the first terms of the sequence $\left((tA_2)^n\right)_{n\in\mathbb N}$, a pattern emerges. Note that\begin{align}(tA_2)^0&=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\\tA_2&=\begin{pmatrix}0&0&0\\0&0&-2t\\0&2t&0\end{pmatrix}\\(tA_2)^2&=\begin{pmatrix}0&0&0\\0&-4t^2&0\\0&0&-4t^2\end{pmatrix}\\(tA_2)^3&=\begin{pmatrix}0&0&0\\0&0&8t^3\\0&-8t^3&0\end{pmatrix}.\end{align}This will allow you to deduce that$$\exp(tA_2)=\begin{pmatrix}1&0&0\\0&\cos(2t)&-\sin(2t)\\0&\sin(2t)&\cos(2t)\end{pmatrix}.$$

Answer 2

A little works shows that $v_1 = (1,1,3)^T, v_2 = (10, 7+i, 8+4i)^T, v_3 = \bar{v_2}$ are linearly independent eigenvectors corresponding to the eigenvalues $\lambda_k$.

With $V=\begin{bmatrix} v_1 & v_2 & v_3 \end{bmatrix}$ that we have

Hence $A = V \operatorname{diag}(\lambda_1, \lambda_k, \lambda_3) V^{-1}$, and so $e^{At} = V \operatorname{diag}(e^{\lambda_1 t}, e^{\lambda_k t}, e^{\lambda_3 t}) V^{-1}$.

If we let $P=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -i \\ 0 & 1 & i \end{bmatrix}$, then $VP$ is real, and we get $R = (VP)^{-1} A (VP) = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix}$, and as before $e^{At} = (VP) e^{Rt} (VP)^{-1}$.

We see that $e^{Rt} = \begin{bmatrix} e^{3t} & 0 \\ 0 & e^{(I+2J)t} \end{bmatrix}$, where $J=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. Since $I,J$ commute, we have $e^{(I+2J)t} = e^{t} e^{2Jt}$, and a small amount of work (using the fact that $J^2 = -I$) shows that $e^{(I+2J)t} = e^{t} (\cos (2t) I + \sin (2t) J)$.

Hence $e^{Rt} = \begin{bmatrix} e^{3t} & 0 & 0 \\ 0 & e^t \cos (2t) & -e^t\sin (2t) \\ 0 & e^t \sin(2t) & e^t \cos(2t)\end{bmatrix}$