📂Matrix Algebra

Positive Semidefinite Matrices and Their Real Powers

Definition

For a positive-definite matrix $A \ge 0$ and a real number $t \in \mathbb{R}$, the $t$-power of $A$ is defined as follows. $$A^{t} := \exp \left( t \log A \right)$$ Here, $\exp$ and $\log$ are respectively the matrix exponential and matrix logarithm.

Explanation

Generally, the power of a matrix $A^{t}$ is defined as taking the product of the matrix $t$ times for a natural number $t \in \mathbb{N}$, and if $A$ is an invertible matrix, there exists an inverse matrix $A^{-1}$ allowing it to be generalized for an integer $t \in \mathbb{Z}$.

Going a step further, if $A$ is a positive-definite matrix, its matrix logarithm $\log A$ exists and since all eigenvalues are positive, $A^{t}$ can be naturally generalized for a real number $t$. The relationship between all eigenvalues being positive and the existence of $A^{t}$ can be readily accepted during the proof of the following theorem.

Theorem

For a positive-definite matrix $A$ and a real number $t$, $A^{t}$ exists uniquely.

Proof

Let’s denote the space of Hermitian matrices and the convex set of positive-definite matrices as $\mathbb{H}_{n}$ and $\mathbb{P}_{n}$, respectively. Note that we must be cautious not to arbitrarily use operations like $e^{t E} = e^{t} e^{E}$ during the proof since $t$ is assumed to be real.

Lemma 1: Unitary Diagonalization of Matrix Exponential

Spectral Theory: It is equivalent for $A$ to be a Hermitian matrix and to be unitarily diagonalizable: $$A = A^{\ast} \iff A = Q \Lambda Q^{\ast}$$

A Hermitian matrix $X \in \mathbb{H}_{n}$ can be diagonalized as $X = Q D Q^{\ast}$ according to Spectral Theory. Here, $Q$ is a unitary matrix, and $D$ is a diagonal matrix consisting of the eigenvalues of $A$. Thus, the following holds: $$\begin{align*} \exp X =& e^{X} \\ =& \sum_{k=0}^{\infty} \frac{1}{k!} X^{k} \\ =& \sum_{k=0}^{\infty} \frac{1}{k!} Q D^{k} Q^{\ast} \\ =& Q \left[ \sum_{k=0}^{\infty} \frac{1}{k!} D^{k} \right] Q^{\ast} \\ =& Q e^{D} Q^{\ast} \end{align*}$$

Lemma 2: Eigenvalues of matrix and Matrix Exponential

If we denote the eigenvalues of $X = Q D Q^{\ast}$ as $d_{1} , \cdots , d_{n}$, that is, if we set $\exp X = Q e^{D} Q^{\ast}$ to be $D = \diag \left( d_{1} , \cdots , d_{n} \right)$, the following holds: $$\begin{align*} e^{X} =& Q \sum_{k=0}^{\infty} {\frac{ 1 }{ k! }} \begin{bmatrix} d_{1} & 0 & 0 \\ 0 & \ddots & 0 \\ 0 & 0 & d_{n} \end{bmatrix}^{k} Q^{\ast} \\ =& Q \begin{bmatrix} \sum_{k} d_{1}^{k} / k! & 0 & 0 \\ 0 & \ddots & 0 \\ 0 & 0 & \sum_{k} d_{n}^{k} / k! \end{bmatrix} Q^{\ast} \\ =& Q \begin{bmatrix} e^{d_{1}} & 0 & 0 \\ 0 & \ddots & 0 \\ 0 & 0 & e^{d_{n}} \end{bmatrix} Q^{\ast} \end{align*}$$ The following necessary and sufficient condition can be derived in the same manner: $$X = Q \begin{bmatrix} d_{1} & 0 & 0 \\ 0 & \ddots & 0 \\ 0 & 0 & d_{n} \end{bmatrix} Q^{\ast} \iff e^{X} = Q \begin{bmatrix} e^{d_{1}} & 0 & 0 \\ 0 & \ddots & 0 \\ 0 & 0 & e^{d_{n}} \end{bmatrix} Q^{\ast}$$

Concrete Form

The matrix logarithm $\log : \mathbb{P}_{n} \to \mathbb{H}_{n}$ of the positive-definite matrix $A$, which is $\log A$, is a Hermitian matrix and thus can be unitarily diagonalized in the same way: $$t \cdot \log A = t \cdot U E U^{\ast} = U \left[ t E \right] U^{\ast}$$ If we consider $\lambda_{1} , \cdots , \lambda_{n}$ to be the eigenvalues of $A$, where $E$ is $\diag \left( \log \lambda_{1}, \cdots, \log \lambda_{n} \right)$, which consists of the logarithms of each eigenvalue arranged in a diagonal matrix, we can observe the concrete form of $A^{t}$. $$\begin{align*} A^{t} =& \exp \left( t \log A \right) \\ =& U e^{t E} U^{\ast} & \because \text{Lemma 1} \\ =& U \begin{bmatrix} e^{t \log \lambda_{1}} & 0 & 0 \\ 0 & \ddots & 0 \\ 0 & 0 & e^{t \log \lambda_{n}} \end{bmatrix} U^{\ast} & \because \text{Lemma 2} \\ =& U \begin{bmatrix} \lambda_{1}^{t} & 0 & 0 \\ 0 & \ddots & 0 \\ 0 & 0 & \lambda_{n}^{t} \end{bmatrix} U^{\ast} \end{align*}$$

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In the proof, the existence of $A^{t}$ is confirmed through Spectral Theory and the fact that the matrix logarithm is bijective, and it’s shown that no other $A^{t}$ can exist since the diagonal matrix neatly presents as a power of positive numbers. For example, the square root matrix $\sqrt{A}$ of a positive-definite matrix $A$ would uniquely exist.

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