행렬의 로그를 계산하자

지수에 행렬을 올리거나, 행렬의 로그를 계산해보자.

여기의 내용은 [1]을 나름대로 재구성해 봤음. (사실 뭐 문제를 그냥 푼 것에 지나지 않지만-_-) [2]를 참고할 것.

0. notation >

$gl(n, C)$ is the set of $n \times n$ matrices whose entries are complex numbers.
$GL(n, C)$ is the subset of $gl(n, C)$ whose determinant is not zero.
$E$ denotes the identity matrix.

1. definition > exp A, log A

$\displaystyle \exp A = E + A + \frac{A^2}{2!} + \cdots + \frac{A^n}{n!}+\cdots$

$\log A$ is a matrix $X$ such that $\exp X = A$.

2. properties > properties of exp A

If $B \in GL(n, C)$ , then $B(\exp A)B^{-1} = \exp(BAB^{-1})$. (easy)
If $AB = BA$ , then $\exp(A + B) = (\exp A)(\exp B)$. (easy)
$\det(\exp A) = \exp(\mathrm{Tr} A)$.
($\because$ Determinent is product of eigenvalues and trace is sum of eigenvalues)
$\therefore \exp A \in GL(n, C)$ for $\forall A \in gl(n, C)$
$(\exp A)^{-1} = \exp(-A)$ . (easy)

3. example > calculation of exp A

Let $A = \begin{bmatrix} 1&2\\3&4 \end{bmatrix}$
The eigenvalues of $A$ are 5, -2. The eignevectors of $A$ are (1, 1), (4, -3).

Let $Q = \begin{bmatrix} 1&4 \\ 1&-3 \end{bmatrix}$ and $D = \begin{bmatrix} 5&0 \\ 0&-2 \end{bmatrix}$

$\therefore Q^{-1}AQ = D \to A = QDQ^{-1}$

$\exp A = \exp(QDQ^{-1}) = Q(\exp D)Q^{-1}$.

Clearly, $\exp D = \begin{bmatrix} e^5 & 0 \\ 0 & e^{-2} \end{bmatrix}$

$\therefore \displaystyle \exp A = \frac{1}{7} \begin{bmatrix} 3e^5 +4e^{-2} & 4e^5 - 4e^{-2} \\ 3e^5 - 3e^{-2} & 4e^5 + 3e^{-2} \end{bmatrix}$

4. fact > Jordan canonical form

Every matrix in $gl(n, C)$ is similar to some Jordan canonical matrix.

5. step 1 > log A

Since the upper fact, $A$ is similar to Jordan canonical matrix. Let $A = QJQ^{-1}$ where $J$ is Jordan canonical matrix. Since $\log A = Q(\log J)Q^{-1}$, it is enough to find $\exp J$.

6. step 2 > log J

Let $A_i$ be an Jordan block matrix of Jordan canonical matrix $J$ . then

$\log J = \begin{bmatrix} \log A_1 & & & \\ & \log A_2 & & \\ & & \cdots & \\ & & & \log A_s \end{bmatrix}$

it is enough to find $\log A_s$ where $A_s$ is Jordan block matrix.

7. step 3 > log As

Write $A_s$ as $(\lambda E) N$ where the diagonal entries of $N$ are 1’s. Let $\lambda E = \exp X_1$ , and $N = \exp X_2$ . Since $X_1$ is diagonal, $X_1 X_2 = X_2 X_1$ .
$\therefore \exp(X_1 + X_2 ) = A_s$ . it is enough to find $\log N$ .

8. step 4 > log N

Observe that since sufficiently high powers of the difference $E - N$ vanishes, $N$ has logarithm.

$\displaystyle \log N = - \sum_{k=1}^{\infty}\frac{(E-N)^k}{k}$

Use formal power series argument to show that $\exp(\log N) = N$ .

9. conclusion > exponential function

Exponential function $\exp A$ for $GL(n, C)$ is surjective.

[1] Frank W. Warner, Foundation of Differential Manifolds and Lie Groups, Springer, Graduate Texts in Mathematics 94, 1983, ISBN 3540908943, p134, Chapter 3, exercise 15
[2] Matrix exponential in Wikipedia