How Can I Compute exp(tA) Using Matrix A?

[Unassuming]

Homework Statement

A=\begin{bmatrix}-1 & 2 & 0\\-2 & -1 & 0\\ 0 & 0 & -3\end{bmatrix}
Use matrix A and compute exp(tA) explicitly.

Homework Equations

I am having trouble figuring out how to start this. I know how to look at each component of matrix A and to use exp(tA)=I+tA +...(exponential expansion) but I am not sure how to simplify each component's expansion. I don't know if there is an easier way to break matrix A up? Please help me to get started!

3.I have noticed that this matrix A is in canonical form with complex eigenvalues and 1 distinct eigenvalue -3. I tried to break this matrix up into 2 matrices (B and C) that commute so that exp(A)=exp(B)exp(C), but matrix C is a nilpotent matrix with all 0's except -3(bottom right corner) and matrix B has an alpha-beta block in the upper right left corner...

The attempt at a solution

 

[ptr]

Notice \begin{bmatrix}-1 & 2 & 0\\-2 & -1 & 0\\ 0 & 0 & -3\end{bmatrix} ^i = \begin{bmatrix}{((-1)^i (2i-1))} & {((-1)^{i-1} 2i)} & 0\\{(-1)^{i} 2i)} & {((-1)^i (2i-1))} & 0\\ 0 & 0 & (-3)^i\end{bmatrix}. The series then is easily represented in a closed form.

 

[Unassuming]

Thank you for your help. Is the closed form the exp(A)? How do I incorporate t?

 

[ptr]

Yes, sorry, I meant, in a series, not closed form, hahah, do you need the actual sum?

 

[Unassuming]

yes please.

 

[ptr]

The actual sum would depend on element of the domain chosen where the image is to be evaluated.

 

[Unassuming]

I checked and noticed that matrix A does not follow component wise the given summation for the series since A^{3}=\begin{bmatrix} 11&-2&0\\2&11&0\\0&0&-27\end{bmatrix} and A^{4}=\begin{bmatrix} -7&24&0\\-24&-7&0\\0&0&81\end{bmatrix}

I have to be able to use exp(tA) to write the system X'=AX in a general solution form. Please help me to figure out how to write exp(tA).

 

[davyjones]

u have to diagonalize the mx first

 

[HallsofIvy]

u have to diagonalize the mx first

mx? There is no "mx" in the problem. If you meant A, the whole point of this problem is that you can't diagonalize A: it is not diagonalizable. Of course, you can separate the -3: obviously e[sups]A[/sup] will have e^-3 3rd row 3rd column and 0s elsewhere on the third row and column.

The whole problem, then, is finding e^B where B is the matrix
\left[\begin{array}{cc}-1 &amp; 2 \\ -2 &amp; -1\end{array}\right][/itex]<br /> That has determinant 5 so we can write it as <br /> 5\left[\begin{array}{cc}-\frac{1}{\sqrt{5}} &amp;amp; \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} &amp;amp; \frac{-1}{\sqrt{5}}\end{array}\right]<br /> and that matrix can be interpreted as a rotation matrix with cos(\theta)= -1/\sqrt{5} and sin(\theta)= 2/\sqrt{5}. (\theta is about 116 degrees but that isn&#039;t important.)

 

[Unassuming]

I think I have part of the solution. I have e^{tA}.

Now I am asked to write the general solution in the form X(t)=e^{tA}X_0,

where X_0=X(0).

I have figured the bulk of the work out, I just don't know exactly what form this is going to look like.

Will it look like this?

\begin{bmatrix}X_1(t) \\ X_2(t) \\ X_3(t) \end{bmatrix} = e^{tA} \begin{bmatrix} X_1(0) \\ X_2(0) \\ X_3(0) \end{bmatrix}.

EDIT: Also, should I plug that 0 in at some time for anything?