How can a projection matrix be used to solve an exponential matrix equation?

[Sigurdsson]

Homework Statement

Show that
$e^{tA} = I - A + e^{t}A$

$t \in T \ \ \ \ T \subset R$

R being the set of real numbers and T some interval.

The matrix A is a projection matrix. i.e. $A^2 = A$

The Attempt at a Solution

First attempt at the problem involved showing that $e^{tA}$ idempotent because of the projection matrix but soon found out that it could not be.

$A^2 = A$ then

$(tA)^2 = tA$ this, I think, is obvious, but then

$(e^{tA})^2 = e^{2tA} \neq e^{tA}$ this is where I stranded first.


Second attempt involved this trick
$e^{tA}e^{-tA} = I$ this will give us

$e^{tA}e^{-tA} = I = Ie^{-ta} - Ae^{-tA} + e^{t-tA}A$

$= (I-A)e^{-tA} + Ae^{t(I-A)}$
Now, if I differentiate this

$\frac{d}{dt} I = 0 = -A(I-A)e^{-tA} + A(I-A)e^{t(I-A)}$

So close, yet so far...this cannot be true cause of the exponential functions. Not unless

$-A = I - A$ by some means.


Any ideas?

 

[micromass]

Cant you just use the definition of the matrix exponential, i.e.

$$e^{tA}=\sum_{k=0}^\infty{\frac{t^kA^k}{k!}}$$

Using the formula A=A² in there should simplify a lot of terms...

 

[Sigurdsson]

You my dear sir are a delight!

$e^{tA} = \sum_{n=0}^{\infty} \frac{1}{n!} t^n A^n = I \ - \ A \ + \ e^t A = I \ - \ A \ + \ A \sum_{n=0}^{\infty} \frac{1}{n!} t^n$

$= I - A + A(I + t + \frac{1}{2!} t^2 \dots \frac{1}{n!}t^n)$

Now we know that $A^n = A$ so

$\sum_{n=0}^{\infty} \frac{1}{n!} t^n A^n$ becomes

$I + tA + \frac{1}{2!} t^2 A + \dots + \frac{1}{n!} t^n A$

This is equivalent to the right side of the equation, so it fits.*victory dance*

 

[micromass]