Matrix exponential

  • Thread starter Unassuming
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  • #1
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Homework Statement



A=[tex]\begin{bmatrix}-1 & 2 & 0\\-2 & -1 & 0\\ 0 & 0 & -3\end{bmatrix}[/tex]
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
 

Answers and Replies

  • #2
ptr
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Notice [tex] \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}[/tex]. The series then is easily represented in a closed form.
 
  • #3
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Thank you for your help. Is the closed form the exp(A)? How do I incorporate t?
 
  • #4
ptr
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Yes, sorry, I meant, in a series, not closed form, hahah, do you need the actual sum?
 
  • #6
ptr
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The actual sum would depend on element of the domain chosen where the image is to be evaluated.
 
  • #7
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I checked and noticed that matrix A does not follow component wise the given summation for the series since A[tex]^{3}[/tex]=[tex]\begin{bmatrix} 11&-2&0\\2&11&0\\0&0&-27\end{bmatrix}[/tex] and [tex]A^{4}[/tex]=[tex]\begin{bmatrix} -7&24&0\\-24&-7&0\\0&0&81\end{bmatrix}[/tex]

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).
 
  • #8
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u have to diagonalize the mx first
 
  • #9
HallsofIvy
Science Advisor
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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 eB where B is the matrix
[tex]\left[\begin{array}{cc}-1 & 2 \\ -2 & -1\end{array}\right][/itex]
That has determinant 5 so we can write it as
[tex]5\left[\begin{array}{cc}-\frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{-1}{\sqrt{5}}\end{array}\right][/tex]
and that matrix can be interpreted as a rotation matrix with [itex]cos(\theta)= -1/\sqrt{5}[/itex] and [itex]sin(\theta)= 2/\sqrt{5}[/itex]. ([itex]\theta[/itex] is about 116 degrees but that isn't important.)
 
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  • #10
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I think I have part of the solution. I have [tex]e^{tA}[/tex].

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

where [tex]X_0=X(0)[/tex].

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?

[tex]\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} [/tex].

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

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