# Matrix exponential relative to eigenvalue

Originally I was having trouble with this linear algebra problem from class but while laying out my work in mathjax I think I found my error and corrected it. Here is my answer although I think there is probably a less roundabout way to get there if anyone wants to post a more concise/precise answer.

Question

let $$A \in {\Re}^{nxn}$$

let $$\lambda$$ be an eigenvalue of $$A$$ with the algebraic multiplicity $$n$$

Prove that for any $$t \in \Re$$

$${e}^{tA} = {e}^{\lambda t} (I+(A-\lambda I)t+...+\frac{(A-\lambda I)^{n-1}}{(n-1)!}t^{n-1})$$

My answer:

I start by representing the exponential in it's Taylor series form because having a polynomial form is easier to work with.

$$e^{t A} = \sum_{i=0}^{\infty} \frac{(t A)^{i}}{i!}$$

which is of the form

$$I+tA+...+\frac{A^{k}}{k!}t^{k}$$

Next I look at the eigenvector denoted by $$v$$ such that

$$Av=\lambda v$$

or equivalently

$$(A-\lambda I)v = 0$$

Applying the eigenvector

$$e^{t A}v = v + t(Av) +...+\frac{(A^{k}v)}{k!} t^{k} = v + t(\lambda v) +...+\frac{\lambda^{k} v}{k!} t^{k}$$

Then pull out the $$v$$ to get the form

$$e^{\lambda t} v$$

so

$$(e^{tA} - e^{\lambda t})v = 0 = ((I+tA+...t^k \frac{A^k}{k!})-(I+(\lambda t) +...+t^{k} \frac{\lambda^{k}}{k!}))v$$

$$(t(A-\lambda I)+...+t^{k} \frac{(A-\lambda I)^{k}}{k!})v = 0$$

It's also clear to me that we only need to examine to $$n-1$$ because at degrees greater and equal to n it is guaranteed to be zero because of the algebraic multiplicity of $$n$$.

$$(t(A-\lambda I)+...+t^{(n-1)} \frac{(A-\lambda I)^{(n-1)}}{(n-1)!})v = 0$$

Plug back into earlier equation gives

$$e^{A t} v = e^{\lambda t}(I+t(A-\lambda I)+...+t^{(n-1)} \frac{(A-\lambda I)^{(n-1)}}{(n-1)!}) v$$

Pull v back out and you get the final equation

$${e}^{tA} = {e}^{\lambda t} (I+(A-\lambda I)t+...+\frac{(A-\lambda I)^{n-1}}{(n-1)!}t^{n-1})$$

## 2 Answers

Your mistake: we have $$e^{t A} = \sum_{i=0}^{\infty} \frac{(t A)^{i}}{i!}$$. In general $$e^{t A} \ne I+tA+...+\frac{A^{k}}{k!}t^{k}$$ !

The characteristic polynomial $$p$$ of $$A$$ has the form $$p(x)=(x- \lambda)^n$$ because $$\lambda$$ has algebraic multiplicity $$n$$.

By Cayley - Hamilton: $$0=p(A)=(A- \lambda I)^n.$$

We have

$$e^{tA}= e^{\lambda t}e^{t(A- \lambda I)}$$.

Now procced and invoke that $$(A- \lambda I)^k=0$$ for all $$k \ge n.$$

• Thanks for the help! – EngineeringStudent Dec 11 '19 at 8:49
• Why is that a mistake? I thought that was just applying the definition of the matrix exponential – llama Dec 11 '19 at 17:07
• in general, the matrix exponential is not a finite sum – Fred Dec 12 '19 at 3:59

I think the easier way is to first notice that $${e}^{tA} = {e}^{\lambda t} (I+(A-\lambda I)t+...+\frac{(A-\lambda I)^{n-1}}{(n-1)!}t^{n-1})$$ Is the same as $$\frac{{e}^{tA}}{{e}^{\lambda t}} = (I+(A-\lambda I)t+...+\frac{(A-\lambda I)^{n-1}}{(n-1)!}t^{n-1})$$ $${e}^{t(A-\lambda I)} = (I+(A-\lambda I)t+...+\frac{(A-\lambda I)^{n-1}}{(n-1)!}t^{n-1})$$ Then proving the rest should be quicker.