Exponent function in Matrix

Algebra Level 4

If A = [ a b − b a ] \displaystyle A= \begin{bmatrix} \;\; a & b\\ -b & a \end{bmatrix} , where a , b a,b are real numbers, then find e A t e^{At} .

[ e a e b e − b e a ] t \displaystyle \begin{bmatrix} e^{a} & e^{b} \\ e^{-b} & e^{a} \end{bmatrix}t e a t [ cos ⁡ b t sin ⁡ b t − sin ⁡ b t cos ⁡ b t ] \displaystyle e^{at}\begin{bmatrix} \cos bt & \sin bt \\ -\sin bt & \cos bt \end{bmatrix} e b t [ cos ⁡ a t sin ⁡ a t − sin ⁡ a t cos ⁡ a t ] \displaystyle e^{bt}\begin{bmatrix} \cos at & \sin at \\ -\sin at & \cos at \end{bmatrix} e b t [ cos ⁡ a t sin ⁡ b t − sin ⁡ b t cos ⁡ a t ] \displaystyle e^{bt}\begin{bmatrix} \cos at & \sin bt \\ -\sin bt & \cos at \end{bmatrix} e b t [ cos ⁡ a t − sin ⁡ a t sin ⁡ a t cos ⁡ a t ] \displaystyle e^{bt}\begin{bmatrix} \cos at & -\sin at \\ \sin at & \cos at \end{bmatrix} e a t [ cos ⁡ a t sin ⁡ b t − sin ⁡ b t cos ⁡ a t ] \displaystyle e^{at}\begin{bmatrix} \cos at & \sin bt \\ -\sin bt & \cos at \end{bmatrix} e a t [ cos ⁡ b t − sin ⁡ b t sin ⁡ b t cos ⁡ b t ] \displaystyle e^{at}\begin{bmatrix} \cos bt & -\sin bt \\ \sin bt & \cos bt \end{bmatrix} [ a t − b t a b t − a b t a t − b t ] \displaystyle \begin{bmatrix} \;\;a^{t}-b^{t} & abt \\ -abt & a^{t}-b^{t} \end{bmatrix}

Uzumaki Nagato Tenshou Uzumaki by uzumaki nagato tenshou u… , 26

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1 solution

Brian Moehring
Feb 11, 2017

Let's start with some basics:

  • If M \mathbf{M} is a square matrix, we define e M = exp ⁡ ( M ) = ∑ k = 0 ∞ M k k ! e^\mathbf{M} = \exp(\mathbf{M}) = \sum_{k=0}^\infty \frac{\mathbf{M}^k}{k!} and note that the infinite series on the right side will always converge component-wise.
  • If M , N \mathbf{M}, \mathbf{N} are square matrices that commute (i.e. M N = N M \mathbf{M}\mathbf{N} = \mathbf{N}\mathbf{M} ), then e M + N = e M e N . e^{\mathbf{M}+\mathbf{N}} = e^\mathbf{M} e^\mathbf{N}.

Therefore, we may write A t = a t [ 1 0 0 1 ] + b t [ 0 1 − 1 0 ] = a t I 2 + b t [ 0 1 − 1 0 ] , At = at\left[\begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right] + bt\left[\begin{array}{cc} 0 & 1 \\ -1 & 0\end{array}\right] = at\mathbf{I}_2 + bt\left[\begin{array}{cc} 0 & 1 \\ -1 & 0\end{array}\right], and since I 2 \mathbf{I}_2 commutes with every matrix, we can use the property I mentioned above to split the problem into two simpler parts:

Therefore, we compute exp ⁡ ( a t I 2 ) = ∑ k = 0 ∞ ( a t ) k I 2 k k ! = ∑ k = 0 ∞ ( a t ) k k ! I 2 = e a t I 2 \exp\left(at\mathbf{I}_2\right) = \sum_{k=0}^\infty \frac{(at)^k\mathbf{I}_2^k}{k!} = \sum_{k=0}^\infty \frac{(at)^k}{k!} \mathbf{I}_2 = e^{at}\mathbf{I}_2 exp ⁡ ( b t [ 0 1 − 1 0 ] ) = ∑ k = 0 ∞ ( b t ) k k ! [ 0 1 − 1 0 ] k = ∑ k = 0 ∞ ( b t ) 2 k ( 2 k ) ! [ 0 1 − 1 0 ] 2 k + ∑ k = 0 ∞ ( b t ) 2 k + 1 ( 2 k + 1 ) ! [ 0 1 − 1 0 ] 2 k + 1 \exp\left(bt\left[\begin{array}{cc} 0 & 1 \\ -1 & 0\end{array}\right]\right) = \sum_{k=0}^\infty \frac{(bt)^k}{k!} \left[\begin{array}{cc} 0 & 1 \\ -1 & 0\end{array}\right]^k = \sum_{k=0}^\infty \frac{(bt)^{2k}}{(2k)!} \left[\begin{array}{cc} 0 & 1 \\ -1 & 0\end{array}\right]^{2k} + \sum_{k=0}^\infty \frac{(bt)^{2k+1}}{(2k+1)!} \left[\begin{array}{cc} 0 & 1 \\ -1 & 0\end{array}\right]^{2k+1}

To simplify this latter expression, note that [ 0 1 − 1 0 ] 2 k = ( − I 2 ) k = ( − 1 ) k I 2 , [ 0 1 − 1 0 ] 2 k + 1 = ( − 1 ) k [ 0 1 − 1 0 ] \left[\begin{array}{cc} 0 & 1 \\ -1 & 0\end{array}\right]^{2k} = (-\mathbf{I}_2)^k = (-1)^k \mathbf{I}_2,\qquad \qquad \left[\begin{array}{cc} 0 & 1 \\ -1 & 0\end{array}\right]^{2k+1} = (-1)^k \left[\begin{array}{cc} 0 & 1 \\ -1 & 0\end{array}\right] so that it becomes exp ⁡ ( b t [ 0 1 − 1 0 ] ) = ∑ k = 0 ∞ ( b t ) 2 k ( − 1 ) k ( 2 k ) ! I 2 + ∑ k = 0 ∞ ( b t ) 2 k + 1 ( − 1 ) k ( 2 k + 1 ) ! [ 0 1 − 1 0 ] = cos ⁡ ( b t ) I 2 + sin ⁡ ( b t ) [ 0 1 − 1 0 ] = [ cos ⁡ ( b t ) sin ⁡ ( b t ) − sin ⁡ ( b t ) cos ⁡ ( b t ) ] \exp\left(bt\left[\begin{array}{cc} 0 & 1 \\ -1 & 0\end{array}\right]\right) = \sum_{k=0}^\infty \frac{(bt)^{2k}(-1)^k}{(2k)!} \mathbf{I}_2 + \sum_{k=0}^\infty \frac{(bt)^{2k+1}(-1)^k}{(2k+1)!} \left[\begin{array}{cc} 0 & 1 \\ -1 & 0\end{array}\right] = \cos(bt)\mathbf{I}_2 + \sin(bt)\left[\begin{array}{cc} 0 & 1 \\ -1 & 0\end{array}\right] = \left[\begin{array}{cc} \cos(bt) & \sin(bt) \\ -\sin(bt) & \cos(bt)\end{array}\right]

Finally, putting it all together, we have e A t = e a t I 2 [ cos ⁡ ( b t ) sin ⁡ ( b t ) − sin ⁡ ( b t ) cos ⁡ ( b t ) ] = e a t [ cos ⁡ ( b t ) sin ⁡ ( b t ) − sin ⁡ ( b t ) cos ⁡ ( b t ) ] . e^{At} = e^{at}\mathbf{I}_2 \left[\begin{array}{cc} \cos(bt) & \sin(bt) \\ -\sin(bt) & \cos(bt)\end{array}\right] = e^{at} \left[\begin{array}{cc} \cos(bt) & \sin(bt) \\ -\sin(bt) & \cos(bt)\end{array}\right].

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution, Brian. Your time in putting all this together is appreciated. There is a more brute force way of demonstrating the same, but this is better in showing some of the interesting properties of matrix exponentiation.

Michael Mendrin - 4 years, 3 months ago

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Thanks! Yeah, there are a lot of ways to see this same result. The fastest way I know of is just to note the eigenvalues of A t At are a t ± b t i at\pm bti , so under a change of basis, e A t e^{At} just looks like e a t ± b t i = e a t ( cos ⁡ ( b t ) ± i sin ⁡ ( b t ) ) e^{at\pm bti}=e^{at}(\cos (bt) \pm i\sin (bt)) . Therefore, the final result can be seen as a real matrix form of complex exponentiation!

Brian Moehring - 4 years, 3 months ago

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I used that approach, but I kinda doubted its validity. Thanks! :)

A Former Brilliant Member - 4 years, 3 months ago

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@A Former Brilliant Member – To be a formal solution, you would need to check a few things (that At is diagonalizable, what the basis-change matrix is), but it is a correct intuition in this case.

Brian Moehring - 4 years, 3 months ago

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@Brian Moehring – Yep, yep. It is a pretty cool method too! :P

A Former Brilliant Member - 4 years, 3 months ago

@SUMUKHA ADIGA

rakshith lokesh - 3 years, 1 month ago

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