📂Matrix Algebra

Logarithm of a Matrix

Introduction¹

Matrix exponential is defined by the following Taylor series. For a matrix $X \in M_{n \times n}(\mathbb{C})$,

$$ e^{X} = \sum\limits_{m=0}^{\infty} \dfrac{X^{m}}{m!} $$

Similarly, the matrix logarithm is also defined by a series. The Taylor series of the logarithm is as follows.

$$ \log z = \sum\limits_{m=1}^{\infty} (-1)^{m+1}\dfrac{(z-1)^{m}}{m}, \qquad |z-1| \lt 1 \tag{1} $$

Definition

$n \times n$ For a matrix $A$, if the following series converges, we define it to be $\log A$.

$$ \log A = \sum\limits_{m=1}^{\infty} (-1)^{m+1} \dfrac{(A - I)^{m}}{m} \tag{2} $$

Remarks

Just as the series $(1)$ converges when $| z - 1 | \lt 1$, the series $\log A$ also converges when $\| A - I \| \lt 1$. Note that this is a sufficient condition, so it does not mean the series converges only when $\| A - I \| \lt 1$. For example, even if $\| A - I \| \gt 1$, the series converges when $A - I$ is a nilpotent matrix.

As with the matrix exponential, if $A \in M_{n \times n}(\mathbb{C})$ then $\log A$ itself is a matrix contained in $M_{n \times n}(\mathbb{C})$.

Be careful about the result in (c): convergence of $\log (e^{X})$ does not guarantee that its value equals $X$. As a counterexample, if we take $X = 2\pi i I$, then $e^{X} = e^{2\pi i} I = I$. Hence $e^{X} - I = O$ and $\log (e^{X}) = O$. In this case, $X \ne \log (e^{X})$.

From (d) we obtain:

$$ \log(I + B) = B + O(\| B^{2} \|) $$

Here $O$ is big-O notation.

Properties

(a) Functions of the form $(2)$ and similar $\log$ are defined on the set $D = \left\{ A \in M_{n \times n}(\mathbb{C}) : \| A - I \| \lt 1 \right\}$ and are continuous functions.

(b) For $A \in D$, $$ e^{\log A} = A $$

　(b’) If $A$ is idempotent, then $e^{\log A} = A$.

(c) For $\| X \| \lt \log 2$ of $X \in M_{n \times n}(\mathbb{C})$, $$ \| e^{X} - I \| \lt 1 \text{ and } \log e^{X} = X $$

　(c’) If $X$ is nilpotent, then $\log( e^{X} ) = X$.

(d) For all $B \in M_{n \times n}(\mathbb{C})$ satisfying $\| B \| \lt 1/2$, there exists a constant $c$ such that the following holds. $$ \| \log (I + B) - B \| \le c \| B \|^{2} $$

Proof

(a)

Properties of convergence of matrices:

$$ \| A_{n} - A \| \to 0 \text{ as } n \to \infty \iff \lim\limits_{n \to \infty} A_{n} = A $$

Expanding the series as sums and limits, we have

$$ \log A = \lim\limits_{M \to \infty} \sum_{m = 1}^{M} (-1)^{m+1} \dfrac{(A - I)^{m}}{m} $$

By the properties of matrix norms, $\| (A - I)^{m} \| \le \| A - I \|^{m}$ holds. Therefore we obtain

$$ \sum\limits_{m = 1}^{M} \left\| (-1)^{m+1} \dfrac{(A - I)^{m}}{m} \right\| \le \sum\limits_{m = 1}^{M} \dfrac{\| A - I \|^{m}}{m} $$

If $\| A - I \| \lt 1$, then by the ratio test the series is absolutely convergent and hence convergent.

Moreover, from the above inequality and the Weierstrass M-test, $\log A$ is continuous.

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(b)

Case 1: $A$ is diagonalizable

Assume $A$ is diagonalizable.

$$ A = C D C^{-1} $$

Then $(A - I)^{m}$ is given by

$$ \begin{align*} (A - I)^{m} &= (CDC^{-1} - CIC^{-1})^{m} \\ &= \left[ C(D - I)C^{-1} \right]^{m} \\ &= C(D - I)^{m}C^{-1} \\ &= C\begin{bmatrix} (d_{1} - 1)^{m} & & 0 \\ & \ddots & \\ 0 & & (d_{n} - 1)^{m} \end{bmatrix}C^{-1} \end{align*} $$

Since $\| A - I \| \lt 1$, for each $i$ we must have $|d_{i} - 1| \lt 1$. Hence we obtain

$$ \begin{align*} \log A &= \sum\limits_{m=1}^{\infty} (-1)^{m+1} \dfrac{(A - I)^{m}}{m} \\ &= \sum\limits_{m=1}^{\infty} \left( C\begin{bmatrix} (-1)^{m+1} \dfrac{(d_{1} - 1)^{m}}{m} & & 0 \\ & \ddots & \\ 0 & & (-1)^{m+1} \dfrac{(d_{n} - 1)^{m}}{m} \end{bmatrix}C^{-1} \right) \\ &= C \left( \sum\limits_{m=1}^{\infty} \begin{bmatrix} (-1)^{m+1} \dfrac{(d_{1} - 1)^{m}}{m} & & 0 \\ & \ddots & \\ 0 & & (-1)^{m+1} \dfrac{(d_{n} - 1)^{m}}{m} \end{bmatrix} \right)C^{-1} \\ &= C \begin{bmatrix} \log d_{1} & & 0 \\ & \ddots & \\ 0 & & \log d_{n} \end{bmatrix}C^{-1} \end{align*} $$

Properties of the matrix exponential:

$$ e^{CXC^{-1}} = Ce^{X}C^{-1} $$

For a diagonal matrix $D = [d_{ii}]$, $$ e^{D} = \begin{bmatrix} e^{d_{11}} & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & e^{d_{nn}} \end{bmatrix} $$

Then by properties of the matrix exponential we have

$$ \begin{align*} e^{\log A} &= C \begin{bmatrix} e^{\log d_{1}} & & 0 \\ & \ddots & \\ 0 & & e^{\log d_{n}} \end{bmatrix}C^{-1} \\ &= C \begin{bmatrix} d_{1} & & 0 \\ & \ddots & \\ 0 & & d_{n} \end{bmatrix}C^{-1} \\ &= CDC^{-1} = A \end{align*} $$

Case 2: $A$ is not diagonalizable

Assume $A$ is not diagonalizable. There exists a sequence of diagonalizable matrices $A_{n}$ converging to $A$.

$$ \lim\limits_{n \to \infty} A_{n} = A, \qquad A_{n} \text{ is diagonalizable} $$

Since the matrix exponential and logarithm are continuous functions, we get

$$ e^{\log A} = e^{ \log (\lim\limits_{n \to \infty} A_{n})} = \lim\limits_{n \to \infty} e^{\log A_{n}} = \lim\limits_{n \to \infty} A_{n} = A $$

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(b')

Proof

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(c)

Since $\| e^{X} - I \| \le e^{\| X \|} - 1$,

$$ \| e^{X} - I \| \le e^{\| X \|} - 1 \le e^{\log 2} - 1 \le 1 $$

Therefore $\log e^{X}$ converges. Showing that $\log e^{X} = X$ holds can be done by the same method as for proving $e^{\log A} = A$.

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(c')

Proof

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(d)

$$ \begin{align*} \log (I + B) - B &= \sum\limits_{m=1}^{\infty} (-1)^{m+1} \dfrac{B^{m}}{m} - B \\ &= \sum\limits_{m=2}^{\infty} (-1)^{m+1} \dfrac{B^{m}}{m} \\ &= B^{2} \sum\limits_{m=2}^{\infty} (-1)^{m+1} \dfrac{B^{m-2}}{m} \end{align*} $$

Since $\| B \| \le 1/2$,

$$ \| \log (I + B) - B \| \le \| B \|^{2} \sum\limits_{m=2}^{\infty} \dfrac{(1/2)^{m-2}}{m} $$

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