Homework Help: Matrix Exponentials

1. Apr 14, 2012

R2Zero

1. The problem statement, all variables and given/known data

Let B =

|1 -2|
|1 3 |

Let C =

|1 4 0|
|0 1 0|
|0 0 2|

Find e$^{B}$ and e$^{C}$

2. Relevant equations

e$^{A}$ = $\Sigma$x$^{n}$/n!

e$^{A}$ = P$^{-1}$e$^{D}$P

3. The attempt at a solution
My professor told me the first step to approaching these types of problems is to find the eigenvalues for both B and C. For B, I get eigenvalues 2+i and 2-i, and I get 1 (double root) and 2 as the eigenvalues for C.

I'm having trouble finding some eigenvectors for B (my teacher did not provide any sufficient or clear examples of eigenstuff involving complex numbers), and I get [1, 0, 0] and [0, 0, 1] as my eigenvectors for C (for e-values 1 and 2, respectively).

I'm sure at this point I'm supposed to construct P, P$^{-1}$ and D for each and somehow use those to find e$^{B}$ and e$^{C}$, but I'm not sure how to go about doing that since my professor did not clearly explain the entire process...

2. Apr 14, 2012

morphism

You should look in a linear book (or online) to see how one typically finds eigenvectors. And while you're at it, look up "diagonalization".

The point of diagonalization here is this: if D is diagonal, say with entries d_1, ..., d_n down the diagonal, then e^D will be diagonal entries $e^{d_1}, \ldots, e^{d_n}$ down the diagonal. This makes the formula you quoted ($A=P^{-1}PD \implies e^A=P^{-1}e^DP$ handy.

Just a tip: the matrix C will turn out to not be diagonalizable. But if you look at it, it's clearly "block" diagonal. So by the same token, it will turn out that e^C will be block diagonal with e^{2x2 block of C} for the first block and e^2 for the second.

So now you have to compute the exp of $\left(\begin{smallmatrix} 1&4\\0&4 \end{smallmatrix}\right)$. For this, I recommend directly using the Taylor series of exp and trying to spot the pattern.

3. Apr 14, 2012

Ray Vickson

You don't need the eigenvectors. It is the case that for any analytical function
$$f(x) = c_0 + c_1 x + c_2 x^2 + \cdots,$$ if we define f(A) for an nxn matrix A as $$f(A) = c_0 I + c_1 A + c_2 A^2 + \cdots,$$ then: if the eigenvalues of A are r1 with multiplicity m1, r2 with multiplicity m2, ... we have: there are matrices E_11,...., E_1m1, E_21,...,E_2m2,... such that
$$f(A) = \sum_{j=1}^{m_1} E_{1j} f^{(j-1)}(r_1) + \sum_{j=1}^{m_2} E_{2j} f^{(j-1)}(r_2) + \cdots,$$
where $f^{0}(r) = f(r), \, f^{(1)}(r) = f^{\prime}(r), \; f^{(2)}(r) = f^{\prime \prime}(r),$ etc. Here, the matrices E_{ij} are independent of the form of the function f.

For the case of C, this says
$f(C) = E_{11}f(1) + E_{12}f'(1) + E_{21} f(2)$ for any f. In particular, if you apply this to f(x) = 1 (f(C) = I), f(x) = x (f(C) = C) and f(x) = x^2 (f(C) = C^2) you can get the matrices $E_{11}, E_{12}, E_{22}.$ Then $e^C = E_{11}e^1 + E_{12} e^1 + E_{22} e^2 = (E_{11} + E_{12}) e^1 + E_{22} e^2.$

RGV