- #1
Sigurdsson
- 25
- 1
Homework Statement
Show that
[itex]e^{tA} = I - A + e^{t}A[/itex]
[itex]t \in T \ \ \ \ T \subset R[/itex]
R being the set of real numbers and T some interval.
The matrix A is a projection matrix. i.e. [itex]A^2 = A[/itex]
The Attempt at a Solution
First attempt at the problem involved showing that [itex]e^{tA}[/itex] idempotent because of the projection matrix but soon found out that it could not be.
[itex]A^2 = A[/itex] then
[itex](tA)^2 = tA[/itex] this, I think, is obvious, but then
[itex](e^{tA})^2 = e^{2tA} \neq e^{tA}[/itex] this is where I stranded first.
Second attempt involved this trick
[itex]e^{tA}e^{-tA} = I[/itex] this will give us
[itex]e^{tA}e^{-tA} = I = Ie^{-ta} - Ae^{-tA} + e^{t-tA}A[/itex]
[itex] = (I-A)e^{-tA} + Ae^{t(I-A)}[/itex]
Now, if I differentiate this
[itex]\frac{d}{dt} I = 0 = -A(I-A)e^{-tA} + A(I-A)e^{t(I-A)}[/itex]
So close, yet so far...this cannot be true cause of the exponential functions. Not unless
[itex]-A = I - A[/itex] by some means.
Any ideas?