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I am asked to find $e^{At}$, where

$A = \begin{bmatrix} 1 & -1 & 1 & 0\\ 1 & 1 & 0 & 1\\ 0 & 0 & 1 & -1\\ 0 & 0 & 1 & 1\\ \end{bmatrix}$.

So let me just find $e^{A}$ for now and I can generalize later.

I notice right away that I can write

$A = \begin{bmatrix} B & I_{2}\\ 0_{22} & B\\ \end{bmatrix}$, where

$B = \begin{bmatrix} 1 & -1\\ 1 & 1\\ \end{bmatrix}$.

I'm sort of making up a method here and I hope it works. Can someone tell me if this is correct?

I write:

$A = \mathrm{diag}(B,B) + \begin{bmatrix}0_{22} & I_{2}\\ 0_{22} & 0_{22}\end{bmatrix}$

Call $S = \mathrm{diag}(B,B)$, and $N = \begin{bmatrix}0_{22} & I_{2}\\ 0_{22} & 0_{22}\end{bmatrix}$

I note that $N^2$ is $0_{44}$, so $e^{N} = \frac{N^{0}}{0!} + \frac{N}{1!} + \frac{N^2}{2!} + \cdots = I_{4} + N + 0_{44} + ... = I_{4} + N$

and that $e^{S} = \mathrm{diag}(e^{B}, e^{B})$ and compute:

$e^{A} = e^{S + N} = e^{S}e^{N} = \mathrm{diag}(e^{B}, e^{B})\cdot[I_{4} + N]$

This reduces the problem to finding $e^B$, which is much easier.

Is my logic correct? I just started writing everything as a block matrix and proceeded as if nothing about the process of finding the exponential of a matrix would change. But I don't really know the theory behind this I'm just guessing how it would work.

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This looks correct to me. –  Pierre-Yves Gaillard Oct 16 '11 at 18:23
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@Kyle, I have just two comments: 1. when you expand the exponential, the LHS should be $e^N$ (not $e^B$) and 2. In your case, $S$ and $N$ commutes, thats why $e^{S+N}=e^Se^N$ works. –  Tapu Oct 16 '11 at 18:38
    
thanks for noticing that mistake. also this is probably a dumb question, but any matrix commutes with a diagonal matrix right? (which would be why $S$ and $N$ are guaranteed to commute)? –  Kyle Schlitt Oct 16 '11 at 18:49
    
In general a diagonal matrix doesn't commute with all matrices. The diagonal entries must also be equal –  Pierre-Yves Gaillard Oct 16 '11 at 18:54
    
Oh ok. Thank you. –  Kyle Schlitt Oct 16 '11 at 20:15
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2 Answers

up vote 2 down vote accepted

A different, but rather specific, strategy would be to use the ring homomorphism $${a+bi\in\mathbb C \mapsto \pmatrix{a&-b \\ b&a}\in\mathbb R^{2\times 2}}$$in the block decomposition. Then your problem is equivalent to finding $$e^{t\pmatrix{1+i & 1\\ 0 & 1+i}}=e^{\pmatrix{t+ti & t\\ 0 & t+ti}}=e^{t+ti}e^{\pmatrix{0&t\\0&0}}=(e^{t+ti})\pmatrix{1&t\\0&1}$$ which unfolds to $$\pmatrix{e^t\cos t & -e^t\sin t & t e^t \cos t & -t e^t \sin t \\ e^t \sin t & e^t \cos t & t e^t \sin t & t e^t \cos t \\ 0 & 0 & e^t\cos t & -e^t\sin t \\ 0&0& e^t\sin t & e^t\cos t }$$

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Consider $M(t) = \exp(t A)$, and as you noticed, it has block-diagonal form $$ M(t) = \left(\begin{array}{cc} \exp(t B) & n(t) \\ 0_{2 \times 2} & \exp(t B) \end{array} \right). $$ Notice that $M^\prime(t) = A \cdot M(t)$, and this results in a the following differential equation for $n(t)$ matrix: $$ n^\prime(t) = \mathbb{I}_{2 \times 2} \cdot \exp(t B) + B \cdot n(t) $$ which translates into $$ \frac{\mathrm{d}}{\mathrm{d} t} \left( \exp(-t B) n(t) \right) = \mathbb{I}_{2 \times 2} $$ which is to say that $n(t) = t \exp(t B)$.

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