Matrix Logarithm
If a square matrix \( A \) is diagonalizable, meaning there exists an invertible matrix \( P \) and a diagonal matrix \( D \) such that: \[ A = P D P^{-1} \] then the logarithm of the matrix can be calculated as: \[ \log(A) = P \log(D) P^{-1} \] where \( \log(D) \) is a diagonal matrix obtained by applying the logarithm to the eigenvalues of \( D \).
When the matrix \( D \) is already diagonal, the computation of the matrix logarithm becomes significantly simpler.
For example, if \( D = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \):
$$ \log(D) = \log \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} = \begin{pmatrix} \log(a) & 0 \\ 0 & \log(b) \end{pmatrix} $$
In general, the matrix logarithm involves finding a matrix that, when exponentiated, returns the original matrix.
For instance, given a matrix \( A \), the matrix logarithm is a matrix \( B \) such that: $$ e^B = A $$ where \( e^B \) is the matrix exponential of \( B \). It’s important to note that the matrix logarithm is not unique because the matrix exponential is not injective. As a result, there can be multiple matrices \( B \) such that \( e^B = A \). The principal logarithm (or principal branch) is defined as the logarithm where the eigenvalues of \( B \) have imaginary parts within the interval \([-π, π]\).
Methods and Conditions for Calculation
The logarithm of a square matrix \( A \) is well-defined but requires specific conditions to be met.
- Invertibility: The matrix \( A \) must be invertible (i.e., its determinant is non-zero).
- Spectrum: All eigenvalues of \( A \) must have positive real parts to ensure the logarithm is properly defined.
Depending on the properties of the matrix, the logarithm can be computed using different methods. Here are the most common approaches:
- Diagonalization
This method is applicable when \( A \) is diagonalizable and its eigenvalues have positive real parts. The diagonalization process includes the following steps:- Diagonalize the matrix \( A \): Compute the eigenvalues and eigenvectors of \( A \), and construct \( P \) (the matrix of eigenvectors) and \( D \) (the diagonal matrix of eigenvalues), such that \( A = P D P^{-1} \).
- Compute the logarithm of the diagonal matrix \( D \): Apply the logarithm to the diagonal elements of \( D \), yielding \( \log(D) \).
- Compute the logarithm of \( A \): Use the relation \( \log(A) = P \log(D) P^{-1} \) to find the logarithm of the original matrix.
- Taylor Series Expansion
If \( A \) is close to the identity matrix \( I \), the logarithm can be approximated using a Taylor series expansion: $$ \log(A) = (A - I) - \frac{(A - I)^2}{2} + \frac{(A - I)^3}{3} - \dots $$ - Jordan Form
When \( A \) is not diagonalizable but has a Jordan form, the logarithm can be computed using \( A = P J P^{-1} \), where \( J \) is the Jordan matrix. This method is more complex and involves matrix polynomials.
A Practical Example
Consider the square matrix \( A \):
$$ A = \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix} $$
Since \( A \) is not diagonal, the first step is to diagonalize it.
To do so, calculate the eigenvalues of \( A \).
The eigenvalues are determined by solving the characteristic equation:
$$ \det(A - \lambda I) = 0 $$
Where \( A - \lambda I = \begin{pmatrix} 1 - \lambda & 0 \\ 2 & 3 - \lambda \end{pmatrix} \).
Compute the determinant:
$$ \det(A - \lambda I) = (1 - \lambda)(3 - \lambda) - 0 \cdot 2 $$
$$ \det(A - \lambda I) = (1 - \lambda)(3 - \lambda) $$
Thus, the characteristic equation becomes:
$$ (1 - \lambda)(3 - \lambda) = 0 $$
The solutions are:
$$ \lambda_1 = 1, \quad \lambda_2 = 3 $$
Next, calculate the eigenvectors for each eigenvalue.
- For \( \lambda_1 = 1 \): Substituting \( \lambda = 1 \) into \( (A - \lambda I) \cdot \begin{pmatrix} x \\ y \end{pmatrix} = 0 \), the system simplifies to...
- For \( \lambda_2 = 3 \): Repeat the process for \( \lambda = 3 \).
With the eigenvectors, construct the matrix \( P \) and the diagonal matrix \( D \):
$$ P = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix} $$
The original matrix \( A \) can now be expressed as:
$$ A = P D P^{-1} $$
Finally, compute the logarithm using:
$$ \log(A) = P \log(D) P^{-1} $$
Since \( D \) is diagonal, \( \log(D) \) is straightforward to compute:
$$ \log(D) = \begin{pmatrix} \log(1) & 0 \\ 0 & \log(3) \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & \log(3) \end{pmatrix} $$
Substituting back, the logarithm of \( A \) becomes:
$$ \log(A) = P \begin{pmatrix} 0 & 0 \\ 0 & \log(3) \end{pmatrix} P^{-1} $$
After performing the matrix multiplication:
$$ \log(A) = \begin{pmatrix} 0 & 0 \\ \log(3) & \log(3) \end{pmatrix} $$
And that’s the logarithm of the matrix \( A \).
