Matrix exponential relative to eigenvalue

Question

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.

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Question

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

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

Prove that for any $t \in \Re$

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

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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

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

Answer 1

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 $k \ge n.$

Answer 2

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.