행렬의 로그를 계산하자

지수에 행렬을 올리거나, 행렬의 로그를 계산해보자.

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

 


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 > matrix exponential, matrix logarithm

\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

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