Properties of the matrix exponential

5.10 To prove 2, we write the definition $$\exp (tX)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}{t}^n{X}^n$$ and then differentiate term-by-term (allowed by part 1 of the Lemma). This gives: $$\frac{d}{dt}\exp (tX)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}n{t}^{n-1}{X}^n.$$ The $n=0$ term vanishes and the $n$ over $n!$ gives $1/(n-1)!$ , so (pulling out a common factor of $X$ ) this equals $$X\sum _{n=1}^{\mathrm{\infty }}\frac{t^{n-1}}{(n-1)!}{X}^{n-1}.$$ Now relabelling $m=n-1$ this becomes $$X\sum _{m=0}^{\mathrm{\infty }}\frac{t^m}{m!}{X}^m=X\exp (tX),$$ as required.

8.15 Let's just illustrate this with an example: take ; then by the calculation from the last video, we get: and this should be equal to $X\exp (tX)$ , which is 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 $\exp (X)\exp (Y)={\sum }_{i=0}^{\mathrm{\infty }}\frac{1}{i!}{X}^i{\sum }_{j=0}^{\mathrm{\infty }}\frac{1}{j!}{Y}^j$ . 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: $$\exp (X)\exp (Y)=\sum _{i=0}^{\mathrm{\infty }}\sum _{j=0}^{\mathrm{\infty }}\frac{1}{i!j!}{X}^i{Y}^j.$$ 12.20 Next, we will write out the double sum as a grid. 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 $$\sum _{k=0}^{\mathrm{\infty }}\sum _{i=0}^k\frac{1}{i!(k-i)!}{X}^i{Y}^{k-i}.$$ (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): $$\sum _{k=0}^{\mathrm{\infty }}\frac{1}{k!}\sum _{i=0}^k\frac{k!}{i!(k-i)!}{X}^i{Y}^{k-i}.$$ The term $\frac{k!}{i!(k-i)!}$ is the binomial coefficient $\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right)$ , so this is equivalent to $$\sum _{k=0}^{\mathrm{\infty }}{(X+Y)}^k$$ provided that $XY=YX$ .

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

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