Finding e^(t*Matrix)

Question

Suppose I have a matrix:

$$\begin{pmatrix}2 & 2 & -2 \\ 5 & 1 & -3 \\ 1 & 5 & -3\end{pmatrix}$$works

I have calculated the eigenvectors and eigenvalues (0). And I want to to find: e^(t*Matrix), where the matrix is listed as above.

The eigenvector is: $$\begin{pmatrix}0.5 \\ 0.5 \\ 1 \end{pmatrix}$$works

I have tried to follow the MIT work on this: [https://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/least-squares-determinants-and-eigenvalues/differential-equations-and-exp-at/MIT18_06SCF11_Ses2.10sol.pdf ] ()

I get the procedure that I will need to calculate: SnS^-1, but I do not understand what n is.

Thank you for any help.

-A

Answer 1

Let $v_1=(1,1,2)$, $v_2=\left(0,\frac14,-\frac14\right)$, and $v_3=\left(\frac18,0,\frac18\right)$. Then $A.v_1=0$, $A.v_2=v_1$, $A.v_3=v_2$, and $\{v_1,v_2,v_3\}$ is a basis of $\mathbb{R}^3$.

Let$$M=\begin{bmatrix}1&0&\frac18\\1&\frac14&0\\2&-\frac14&\frac18\end{bmatrix};$$the columns of $M$ are the vectors $v_1$, $v_2$, and $v_3$. Then, if $A$ is your matrix,$$M^{-1}AM=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}$$and therefore\begin{align}M^{-1}e^{tA}M&=\exp\left(\begin{bmatrix}0&t&0\\0&0&t\\0&0&0\end{bmatrix}\right)\\&=\begin{bmatrix}1&t&\frac{t^2}2\\0&1&t\\0&0&1\end{bmatrix}.\end{align}So,\begin{align}e^{tA}&=M\begin{bmatrix}1&t&\frac{t^2}2\\0&1&t\\0&0&1\end{bmatrix}M^{-1}\\&=\begin{bmatrix}6 t^2+2 t+1 & -2 (t-1) t & -2 t (t+1) \\ t (6 t+5) & -2 t^2+t+1 & -t (2 t+3) \\ t (12 t+1) & (5-4 t) t & -4 t^2-3 t+1\end{bmatrix}\end{align}

Answer 2

There’s no need to go to all the effort of computing a Jordan decomposition of this matrix (call it $A$). Since its only eigenvalue is zero, it is nilpotent, and so the power series for $e^{tA}$ consists of a finite number of terms—at most three, in fact, because the characteristic polynomial of $A$ is $\lambda^3$. It’s a fairly straightforward matter to then compute $$e^{tA} = I+tA+\frac12t^2A^2 = \begin{bmatrix} 6t^2+2t+1 & -2t^2+2t & -2t^2-2t \\ 6t^2+5t & -2t^2+t+1 & -2t^2-3t \\ 12t^2+t & -4t^2+5t & -4t^2-3t+1 \end{bmatrix}.$$