Properties of the matrix exponential

Proof:

5.10 To prove 2, we write the definition

$\exp(tX)=\sum_{n=0}^{\infty}\frac{1}{n!}t^{n}X^{n}$ and then differentiate term-by-term (allowed by part 1 of the Lemma). This gives:

$\frac{d}{dt}\exp(tX)=\sum_{n=0}^{\infty}\frac{1}{n!}nt^{n-1}X^{n}.$ The

$n=0$ term vanishes and the

$n$ over

$n!$ gives

$1/(n-1)!$ , so (pulling out a common factor of

$X$ ) this equals

$X\sum_{n=1}^{\infty}\frac{t^{n-1}}{(n-1)!}X^{n-1}.$ Now relabelling

$m=n-1$ this becomes

$X\sum_{m=0}^{\infty}\frac{t^{m}}{m!}X^{m}=X\exp(tX),$ as required.

8.15 Let's just illustrate this with an example: take $X=\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}$ ; then by the calculation from the last video, we get:

$\frac{d}{dt}\exp\left(t\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}\right)=\frac{d}{dt}\begin{pmatrix}\cos t&-\sin t\\ \sin t&\cos t\end{pmatrix}=\begin{pmatrix}-\sin t&-\cos t\\ \cos t&-\sin t\end{pmatrix}$ and this should be equal to

$X\exp(tX)$ , which is

$\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}\begin{pmatrix}\cos t&-\sin t\\ \sin t&\cos t\end{pmatrix}=\begin{pmatrix}-\sin t&-\cos t\\ \cos t&-\sin t\end{pmatrix}$ as expected.

10.00 (4) follows from (3) by taking $Y=-X$ : since $X$ commutes with $-X$ , we get $\exp(X)\exp(-X)=\exp(X-X)=\exp(0)=I$ .

11.10 Finally, we will prove (3). Let's write out the product $\exp(X)\exp(Y)=\sum_{i=0}^{\infty}\frac{1}{i!}X^{i}\sum_{j=0}^{\infty}\frac{1}% {j!}Y^{j}$ . We will do various manipulations which are justified by the absolute convergence claimed in part (1) of the Lemma. First, let's take all the summations outside:

$\exp(X)\exp(Y)=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\frac{1}{i!j!}X^{i}Y^{j}.$ 12.20 Next, we will write out the double sum as a grid.

$\begin{array}[]{c|cccc}.&i=0&i=1&i=2&\cdots\\ \hline j=0&I&X&\frac{1}{2}X^{2}&\cdot^{\cdot^{\cdot}}\\ j=1&Y&XY&\cdot^{\cdot^{\cdot}}&.\\ j=2&\frac{1}{2}Y^{2}&\cdot^{\cdot^{\cdot}}&.&.\\ \vdots&\cdot^{\cdot^{\cdot}}&.&.&.\end{array}$ 13.40 Let's group together the constant term

$I$ , the linear terms

$X+Y$ , the quadratic terms

$\frac{1}{2}X^{2}+XY+\frac{1}{2}Y^{2}$ , etc. Let's call these groups the

$k=0$ ,

$k=1$ ,

$k=2$ terms etc (so

$k$ is the order of the terms in question).

14.30 With this regrouping of terms, our infinite sum becomes

$\sum_{k=0}^{\infty}\sum_{i=0}^{k}\frac{1}{i!(k-i)!}X^{i}Y^{k-i}.$ (Here, in each group,

$j$ is determined by

$i$ as

$j=k-i$ .)

16.10 The inner sum is a finite sum. It looks a lot like a binomial expansion, but not quite. To make it look more like a binomial expansion, we multiply the sum by $k!$ and simultaneously divide it by $k!$ (leaving the answer unchanged):

$\sum_{k=0}^{\infty}\frac{1}{k!}\sum_{i=0}^{k}\frac{k!}{i!(k-i)!}X^{i}Y^{k-i}.$ The term

$\frac{k!}{i!(k-i)!}$ is the binomial coefficient

$\left(\begin{array}[]{c}k\\ i\end{array}\right)$ , so this is equivalent to

$\sum_{k=0}^{\infty}(X+Y)^{k}$ provided that $XY=YX$ .

To finish the proof, note that $\sum_{k=0}^{\infty}\frac{1}{k!}(X+Y)^{k}$ equals $\exp(X+Y)$ by inspection.

17.30 We really need $XY=YX$ . For example, $(X+Y)^{2}=X^{2}+XY+YX+Y^{2}\neq X^{2}+2XY+Y^{2}$ , unless $XY=YX$ .

Remark:

19.10 The double-sum rearrangement in the proof of 4 is called the Cauchy product formula, and works whenever you have absolutely convergent power series.

Properties of the matrix exponential

Proof of the lemma