E Raised to the Power of a Matrix

[kq6up]

Homework Statement

Use a spectrally decomposed matrix and the power series to yield $$e^{M}$$ where $$M=\begin{pmatrix} 5 & -2 \\ -2 & 2 \end{pmatrix}$$.

Homework Equations

$${ M }^{ n }=C{ D }^{ n }{ C }^{ -1 }$$

The Attempt at a Solution

Is this correct? $${ e }^{ M }=\frac { 1 }{ 5 } \begin{pmatrix} 1 & 1 \\ 2 & \frac { -1 }{ 2 } \end{pmatrix}\left( \sum _{ i=0 }^{ \infty }{ \frac { \begin{pmatrix} { 1 }^{ n } & 0 \\ 0 & { 6 }^{ n } \end{pmatrix} }{ n! } } \right) \begin{pmatrix} 1 & 2 \\ 4 & -2 \end{pmatrix}$$

Thanks,
Chris

 

[D H]

Yes, that's correct, but you really should expand that out.

 

[kq6up]

Is this better? $$CI{ C }^{ -1 }+CD{ C }^{ -1 }+\frac { 1 }{ 2 } C{ D }^{ 2 }{ C }^{ -1 }+\frac { 1 }{ 6 } C{ D }^{ 3 }{ C }^{ -1 }\dots$$.

Chris

 

[D H]

No. It's a 2x2 matrix. There should be four simple expressions, one for each element of the matrix.

 

[kq6up]

There might be some small error in there because when I use == with $$Ce^{D}C^-1$$ in Sage, I get a false. I am taking that to mean they are not equivalent.

$${ e }^{ 7/2 }\left( \begin{array}{rr} \frac { 1 }{ 2 } \, \cosh \left( \frac { 5 }{ 2 } \right) +\frac { 3 }{ 10 } \, \sinh \left( \frac { 5 }{ 2 } \right) & -\frac { 2 }{ 5 } \, \sinh \left( \frac { 5 }{ 2 } \right) \\ -\frac { 2 }{ 5 } \, \sinh \left( \frac { 5 }{ 2 } \right) & \frac { 1 }{ 2 } \, \cosh \left( \frac { 5 }{ 2 } \right) -\frac { 3 }{ 10 } \, \sinh \left( \frac { 5 }{ 2 } \right) \end{array} \right)$$

Thanks,
Chris Maness

 

[D H]

You should be able to compute $\exp(D)$ by hand. You do not need a symbolic math processor for this. You don't need a symbolic math processor for any of this. It's a product of three 2x2 matrices.

 

[kq6up]

I did it by hand. I was just checking my results on Sage.

Chris

 

[D H]

What did you do by hand? That matrix in post #5? How did you arrive at that result?

What did you calculate $\exp(D)$ to be?

 

[kq6up]

There was another problem in the set that was similar, and I had the answer. So I knew that it would have sinh(x), cosh(x) in the answer because of this answer, so I just made it fit that form so I can use sinh(x) and cosh(x). It didn't come out as pretty I was hoping for.

Thanks,
Chris Maness

 

[D H]

Yes, you can express this result using the hyperbolic functions. You made an error somewhere, however.

If you solve the problem as suggested you will *not* be using hyperbolic functions. What did you get for $\exp(D)$? Did you even try doing that? That is part of what the question asks you to find. It is very simple result.

 

[kq6up]

Yes, exp(D) is $$\begin{pmatrix} { e }^{ 1 } & 0 \\ 0 & { e }^{ 6 } \end{pmatrix}$$.

Chris

 

[kq6up]

This is what I got before I tried to convert it to hyperbolic functions:

$$\left( \begin{array}{rr} \frac { 4 }{ 5 } \, e^{ 6 }+\frac { 1 }{ 5 } \, e & -\frac { 2 }{ 5 } \, e^{ 6 }+\frac { 2 }{ 5 } \, e \\ -\frac { 2 }{ 5 } \, e^{ 6 }+\frac { 2 }{ 5 } \, e & \frac { 1 }{ 5 } \, e^{ 6 }+\frac { 4 }{ 5 } \, e \end{array} \right) $$

I did it by hand and with Sage. I got the same result.

Thanks,
Chris Maness

 

[D H]

That's the correct result. You made some error in converting that to hyperbolic expressions.

 

[Ray Vickson]

This is what I got before I tried to convert it to hyperbolic functions:

$$\left( \begin{array}{rr} \frac { 4 }{ 5 } \, e^{ 6 }+\frac { 1 }{ 5 } \, e & -\frac { 2 }{ 5 } \, e^{ 6 }+\frac { 2 }{ 5 } \, e \\ -\frac { 2 }{ 5 } \, e^{ 6 }+\frac { 2 }{ 5 } \, e & \frac { 1 }{ 5 } \, e^{ 6 }+\frac { 4 }{ 5 } \, e \end{array} \right) $$

I did it by hand and with Sage. I got the same result.

Thanks,
Chris Maness

Why would you want to try to convert to hyperbolic functions? It would just make everything longer and messier. Isn't it easier to just write $e$ instead of $\sinh(1)+\cosh(1)?$