POSITIVE SEMI-DEFINITE MATRICES

Alicia Y. Tsai; Authors: Laurent EI Ghaoui; Contributors: Phan Hoang, Tony Tin, Ha To Tam, Nguyen Van Kep, Le Dinh Nam, Juliette Decugis, Nguyen Quoc Cuong.; Giuseppe C. Calafiore

30 POSITIVE SEMI-DEFINITE MATRICES

30.1. Definitions

For a given symmetric matrix $A\in \mathbb{R}^{n\times n}$ , the associated quadratic form is the function $q: \mathbb{R}^n \rightarrow \mathbb{R}$ with values

$\begin{align*}q(x) = x^TAx.\end{align*}$

A symmetric matrix $A$ is said to be positive semi-definite (PSD, notation: $A \succeq 0$ ) if and only if the associated quadratic form $q$ is non-negative everywhere:

$\begin{align*}q(x) \ge 0 \quad (\forall x\in \mathbb{R}^n).\end{align*}$

It is said to be positive definite (PD, notation: $A \succ 0$ ) if the quadratic form is non-negative and definite, that is, $q(x) = 0$ if and only if $x=0$ .

It turns out that a matrix is PSD if and only if the eigenvalues of $A$ are non-negative. Thus, we can check if a form is PSD by computing the eigenvalue decomposition of the underlying symmetric matrix.

Theorem: eigenvalues of PSD matrices

A quadratic form $q(x)= x^TAx$ , with $A \in {\bf S}^n$ is non-negative (resp. positive-definite) if and only if every eigenvalue of the symmetric matrix $A$ is non-negative (resp. positive).

Proof.

By definition, the PSD and PD properties are properties of the eigenvalues of the matrix only, not of the eigenvectors. Also, if the $n \times n$ matrix $A$ is PSD, then for every matrix $B$ with $n$ columns, the matrix $B^TAB$ also is.

30.2. Special cases and examples

Symmetric dyads

Special cases of PSD matrices include symmetric dyads. Indeed, if $A = uu^T$ for some vector $u \in \mathbb{R}^n$ , then for every $x$ :

$\begin{align*}q_A(x) = x^Tuu^Tx = (u^Tx)^2 \ge 0.\end{align*}$

More generally if $B \in \mathbb{R}^{m\times n}$ , then $A = B^TB$ is PSD, since

$\begin{align*}q_A(x) = x^TB^TBx = ||Bx||_2^2 \ge 0.\end{align*}$

Diagonal matrices

A diagonal matrix is PSD (resp. PD) if and only if all of its (diagonal) elements are non-negative (resp. positive).

Examples of PSD matrices

30.3. Square root and Cholesky decomposition

For PD matrices, we can generalize the notion of the ordinary square root of a non-negative number. Indeed, if $A$ is PSD, there exists a unique PSD matrix, denoted $A^{1/2}$ , such that $A = (A^{1/2})^2$ . We can express this matrix square root in terms of the SED of $A = U \Lambda U^T$ , as $A^{1/2} = U \Lambda^{1/2} U^T$ , where $\Lambda^{1/2}$ is obtained from $\Lambda$ by taking the square root of its diagonal elements. If $A$ is PD, then so is its square root.

Any PSD matrix can be written as a product $A= LL^T$ for an appropriate matrix $L$ . The decomposition is not unique, and $L = A^{1/2}$ is only a possible choice (the only PSD one). Another choice, in terms of the SED of $A = U^T \Lambda U$ , is $L = U^T \Lambda^{1/2}$ . If $A$ is positive-definite, then we can choose $L$ to be lower triangular, and invertible. The decomposition is then known as the Cholesky decomposition of $A$ .

30.4. Ellipsoids

There is a strong correspondence between ellipsoids and PSD matrices.

Definition

We define an ellipsoid to be an affine transformation of the unit ball for the Euclidean norm:

$\begin{align*}{\bf E} = \{\hat{x} +Lz: ||z||_2 \leq 1 \},\end{align*}$

where $L \in \mathbb{R}^{n\times n}$ is an arbitrary non-singular matrix. We can express the ellipsoid as

$\begin{align*} \mathbf{E}=\left\{x:\left\|L^{-1}(x-\hat{x})\right\|_2 \leq 1\right\}=\left\{x:(x-\hat{x})^T A^{-1}(x-\hat{x}) \leq 1\right\},\end{align*}$

where $A: = L^{-T} L^{-1}$ is PD.

Geometric interpretation via SED

We can interpret the eigenvectors and associated eigenvalues of $A$ in terms of the geometrical properties of the ellipsoid, as follows. Consider the SED of $A$ : $A = U \Lambda U^T$ , with $U^T U = I$ and $\Lambda$ diagonal, with diagonal elements positive. The SED of its inverse is $A^{-1} = LL^T = U \Lambda^{-1} U^T$ . Let $\tilde{x} = U^T(x- \hat{x})$ . We can express the condition $x \in {\bf E}$ as

$\begin{align*}\tilde{x}^T\Lambda \tilde{x} = \sum\limits_{i=1}^n \lambda_i \tilde{x}_i^2 \leq 1.\end{align*}$

Now set $\overline{x}_i = \sqrt{\lambda_i} \tilde{x}_i$ , $i = 1, \cdots, n$ . The above can be written as $\overline{x}^T \overline{x} \leq 1$ : in $\overline{x}$ -space, the ellipsoid is simply a unit ball. In $\tilde{x}$ -space, the ellipsoid corresponds to scaling each $\overline{x}$ -axis by the square roots of the eigenvalues. The ellipsoid has principal axes parallel to the coordinate axes in $\tilde{x}$ -space. We then apply a rotation and a translation to get the ellipsoid in the original $x$ -space. The rotation is determined by the eigenvectors of $A^{-1}$ , which are contained in the orthogonal matrix $U$ . Thus, the geometry of the ellipsoid can be read from the SED of the PD matrix $A^{-1} = LL^T$ : the eigenvectors give the principal directions, and the semi-axis lengths are the square root of the eigenvalues.

	The graph on the left shows the ellipsoid $\{\hat{x} +Lz: \|\|z\|\|_2 \leq 1 \}$ , with
	$\begin{align}\hat{x}=\left(\begin{array}{l} 2 \\ 1 \end{array}\right), \quad L=\left(\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right) \text {. }\end{align}$
	The matrix $LL^T$ admits the SED
	$\begin{align} L L^T &= U^T \Lambda U, \end{align}$
	with
	$\begin{align} U &= \left(\begin{array}{cc} -0.9871 & 0.1602 \\ 0.1602 & 0.9871 \end{array}\right), \\ \Lambda &= \left(\begin{array}{cc} 0.6754 & 0 \\ 0 & 13.3246 \end{array}\right). \end{align}$
	We check that the columns of $U$ determine the principal directions, and $\sqrt{\lambda_1} = 3.6503$ , $\sqrt{\lambda_2} = 0.8219$ are the semi-axis lengths.

The above shows in particular that an equivalent representation of an ellipsoid is

$\begin{align*}\{x: (x-\hat{x})^TB(x-\hat{x}) \leq 1 \}\end{align*}$

where $B:=A^{-1}$ is PD.

It is possible to define degenerate ellipsoids, which correspond to cases when the matrix $B$ in the above, or its inverse $A$ , is degenerate. For example, cylinders or slabs (intersection of two parallel half-spaces) are degenerate ellipsoids.

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	The graph on the left shows the ellipsoid $\{\hat{x} +Lz: \|\|z\|\|_2 \leq 1 \}$ , with
	$\begin{align}\hat{x}=\left(\begin{array}{l} 2 \\ 1 \end{array}\right), \quad L=\left(\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right) \text {. }\end{align}$
	The matrix $LL^T$ admits the SED
	$\begin{align} L L^T &= U^T \Lambda U, \end{align}$
	with
	$\begin{align} U &= \left(\begin{array}{cc} -0.9871 & 0.1602 \\ 0.1602 & 0.9871 \end{array}\right), \\ \Lambda &= \left(\begin{array}{cc} 0.6754 & 0 \\ 0 & 13.3246 \end{array}\right). \end{align}$
	We check that the columns of $U$ determine the principal directions, and $\sqrt{\lambda_1} = 3.6503$ , $\sqrt{\lambda_2} = 0.8219$ are the semi-axis lengths.