# Math Help - Exponential Matrix

1. ## Exponential Matrix

Find $e^{At}$ of : $\begin{pmatrix}0&2&1\\-1&-3&-1\\1&1&-1\end{pmatrix}$ . I have found $\lambda=2, i, -i$ but I get stuck trying to find e-vectors and so on. Thanks.

2. Hello,
Originally Posted by Chief65
Find $e^{At}$ of : $\begin{pmatrix}0&2&1\\-1&-3&-1\\1&1&-1\end{pmatrix}$ . I have found $\lambda=2, i, -i$ but I get stuck trying to find e-vectors and so on. Thanks.
I find the characteristic polynomial : $-(\lambda+1)^2(\lambda+2)$

Maybe this would simplify your research of the eigenvectors

3. Originally Posted by Moo
Hello,

I find the characteristic polynomial : $-(\lambda+1)^2(\lambda+2)$

Maybe this would simplify your research of the eigenvectors
D'oh! Mixed up problems as I was writing them down. The actual matrix is $\begin{pmatrix}2&0&1\\1&0&1\\1&-2&0\end{pmatrix}$ with $\lambda=2, i, -i$

4. Originally Posted by Chief65
D'oh! Mixed up problems as I was writing them down. The actual matrix is $\begin{pmatrix}2&0&1\\1&0&1\\1&-2&0\end{pmatrix}$ with $\lambda=2, i, -i$
Okay

Then, you have to find the vectors $\bold{x}_i=\begin{pmatrix} a \\ b \\ c \end{pmatrix}$

Such that $A\bold{x}_i=\lambda_i \bold{x}_i$

$\begin{pmatrix}2&0&1\\1&0&1\\1&-2&0\end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix}=\begin{pmatrix} 2a+c \\ a+c \\ a-2b \end{pmatrix}$

Let's do it for $\lambda=i$ (for $\lambda=2$, it's common techniques !)

$\begin{pmatrix} 2a+c \\ a+c \\ a-2b \end{pmatrix}=i \begin{pmatrix} a \\ b \\ c \end{pmatrix}=\begin{pmatrix} ai \\ bi \\ ci \end{pmatrix}$

That is :
$\begin{cases} 2a+c=ai \quad (1)\\ a+c=bi \quad (2)\\ a-2b=ci \quad (3)\end{cases}$

From (2), you get $b=\frac{a+c}{i}=-i (a+c)$
Substitute in (3) :
$a+2i(a+c)=ci$
$a(1+2i)+ci=0$

Now let for example a=1 (that's what's good with eigenvectors, you can choose arbitrarily one component)
So $1+2i+ci=0 \Rightarrow c=-\frac{1+2i}{i}=i-2$

And now compute $b=-i(a+c)$

that's all

Similar one for $\lambda=-i$
For $\lambda=2$, it should be easy, because all the coefficients are real.

Once you get the 3 eigenvectors, place them in columns to form matrix $P=[x_1,x_2,x_3]$ such that $A=PDP^{-1}$
and where $D=\begin{pmatrix} \lambda_1 &0&0 \\ 0&\lambda_2 &0 \\ 0&0&\lambda_3 \end{pmatrix}$
(the order of the eigenvalues has to be the same as the order of the eigenvectors in P)

5. Just want to confirm my thought process on backsolving for A given an exponential matrix $e^{At}$.

A matrix of sums is also a sum of matrices, and vice versa. Right? For example,
would it go like this?
$e^{At} = I + \sum_{n=1}^\infty \frac{A^nt^n}{n!} = I + \sum_{n=1}^\infty\begin{pmatrix}a_{11}&a_{12}&a_{1 3}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end {pmatrix}^n\frac{t^n}{n!}$

$
= I + \begin{pmatrix}\sum_{n=1}^\infty \frac{(a_{11}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{12}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{13}t)^n}{n!}\\ \sum_{n=1}^\infty \frac{(a_{21}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{22}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{23}t)^n}{n!}\\\sum_{n=1}^\infty \frac{(a_{31}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{32}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{33}t)^n}{n!}\end{pmatrix}
$

Once I find my $A^n$, just plug in n=1 to get A. Right?

6. Originally Posted by seadog
Just want to confirm my thought process on backsolving for A given an exponential matrix $e^{At}$.

A matrix of sums is also a sum of matrices, and vice versa. Right? For example,
would it go like this?
$e^{At} = I + \sum_{n=1}^\infty \frac{A^nt^n}{n!} = I + \sum_{n=1}^\infty\begin{pmatrix}a_{11}&a_{12}&a_{1 3}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end {pmatrix}^n\frac{t^n}{n!}$

$
= I + \begin{pmatrix}\sum_{n=1}^\infty \frac{(a_{11}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{12}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{13}t)^n}{n!}\\ \sum_{n=1}^\infty \frac{(a_{21}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{22}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{23}t)^n}{n!}\\\sum_{n=1}^\infty \frac{(a_{31}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{32}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{33}t)^n}{n!}\end{pmatrix}
$

Once I find my $A^n$, just plug in n=1 to get A. Right?
Nope.

The n-th power of a matrix is not the matrix formed by the elements to the n-th power ~

Try $A=\begin{pmatrix} 2&3 \\ 3&4 \end{pmatrix}$
Do you have $A^2=\begin{pmatrix} 2^2 & 3^2 \\ 3^2 & 4^2 \end{pmatrix}$ ?
You will see that it's not the case.

You don't need to find A, you need to find an expression for $A^n$
And by finding its eigenvalues and the transition matrix, P, you'll have :
$A=PDP^{-1}$
And hence $A^n=PDP^{-1}PDP^{-1}\dots PDP^{-1}PDP^{-1}=P\underbrace{D\dots D}_{n ~times} P^{-1}$

But as I said above, the n-th power of a diagonal matrix is a matrix formed by the n-th power of its diagonal elements.

So finally, you have :
$e^{At}=I+\sum_{n \geq 1} \frac{A^n t^n}{n!}=I+P \sum_{n\geq 1} \frac{D^nt^n}{n!} P^{-1}=I+P\begin{pmatrix} \sum_{n\geq 1} \frac{\lambda_1^n t^n}{n!} & 0 \\ 0 & \sum_{n\geq 1} \frac{\lambda_2^n t^n}{n!} \end{pmatrix} P^{-1}$

Does it look clear ?