Show that : $e^{tA}=I+g(t,a)xy^T$

Question

Suppose $x,y \in \Bbb R^n$ and let $A = xy^T$. Futher, let $\alpha = x^Ty$. Show That $e^{tA} = I + g(t,\alpha)xy^T$ where $$g(t,\alpha)=\begin{cases}{1\over\alpha}(e^{\alpha t}-1)&\text{if $\alpha \neq 0$}\\t &\text {if $\alpha=0$}\end{cases}$$

I can proof with condition $\alpha = 0$. But I cant solve when $\alpha \neq 0$

When $\alpha = 0$. We need to show that $$e^{tA}=I+txy^T=I+tA$$ We have : $$e^{tA} = \sum_{k=0}^\infty{{t^kA^k}\over k!}$$ Note that : $$AA=A^2=xy^Txy^T=yx^Tyx^T=y\alpha x^T $$ $$AAA=A^3=y\alpha x^Tyx^T=y\alpha \alpha x^T = y\alpha^2x^T$$ So : $$A^k=y\alpha^{k-1}x^T$$ then : $\alpha = 0 , A^k = 0 \text { with $k > 1$}$ $\Rightarrow e^{tA} = \sum_{k=0}^\infty{{t^kA^k}\over k!}=I+tA$

Answer 1

Observe that $\alpha^k$ is a number, and commutes with matrices. So, if $\alpha \neq 0$, we have

$$ e^{tA} = \sum\limits_{k=0}^\infty \frac{A^kt^k}{k!} = \sum\limits_{k=0}^\infty\frac{\alpha^{k-1}t^k}{k!}xy^T = \frac{1}{\alpha}xy^T\sum\limits_{k=0}^\infty\frac{\alpha^kt^k}{k!} = \frac{1}{\alpha}xy^Te^{\alpha t} $$

I hope you can go from here.

Answer 2

There's something wrong with our computation of the powers of $A$. If $x$ and $y$ are not colinear, then $xy^{T}\neq yx^{T}$. On the other hand, $x^{T}y=y^{T}x$ always hold. So you get $$A^2=AA=xy^{T}xy^{T}=\alpha xy^{T}=\alpha A.$$ By induction, we get that $A^{n+1}=\alpha^nA$ for all $n\geq 0$. Thus we have $$e^{tA} = \sum_{k=0}^\infty\frac{t^kA^k}{k!} = I+ \left(\sum_{k=1}^\infty \frac{t^k\alpha^{k-1}}{k!}\right)A,$$ So all that remains to do is to prove that $$\sum_{k=1}^\infty \frac{t^k\alpha^{k-1}}{k!}=g(t,\alpha).$$ You have already solved it for $\alpha=0$; when $\alpha\neq 0$, we get $$\sum_{k=1}^\infty \frac{t^k\alpha^{k-1}}{k!} = \frac{1}{\alpha}\sum_{k=1}^\infty \frac{t^k\alpha^{k}}{k!}=\frac{1}{\alpha}\left(\sum_{k=0}^\infty \frac{(t\alpha)^{k}}{k!}-1\right)=\frac{e^{t\alpha}-1}{\alpha}.$$