# Intuitive explanation of a positive semidefinite matrix

One intuitive definition is as follows. Multiply any vector with a positive semi-definite matrix. The angle between the original vector and the resultant vector will always be less than or equal $\frac{\pi}{2}$. The positive definite matrix tries to keep the vector within a certain half space containing the vector. This is analogous to what a positive number does to a real variable. Multiply it and it only stretches or contracts the number but never reflects it about the origin.

First I'll tell you how I think about Hermitian positive-definite matrices. A Hermitian positive-definite matrix $M$ defines a sesquilinear inner product $\langle Mv, w \rangle = \langle v, Mw \rangle$, and in fact every inner product on a finite-dimensional inner product space $V$ has this form. In other words it is a way of computing angles between vectors, or a way of projecting vectors onto other vectors; over the real numbers it is the key ingredient to doing Euclidean geometry. An inner product can be recovered from the norm $\langle Mv, v \rangle = \langle v, Mv \rangle$ it induces, and a norm in turn can be recovered from its unit sphere $\{ v : \langle Mv, v \rangle = 1 \}$. This unit sphere is a distorted version of the usual unit sphere; the distortions will occur along axes corresponding to the eigenvectors of $M$, and the amount of distortion corresponds to the (inverses of the) corresponding eigenvalues. For example when $\dim V = 2$ it is an ellipse and when $\dim V = 3$ it is an ellipsoid.

A Hermitian positive-semidefinite matrix $M$ no longer describes an inner product because it is not necessarily positive-definite, but it still defines a sesquilinear form. It also defines a function $\langle Mv, v \rangle$ which is no longer a norm because it is not necessarily positive-definite; some people call these "pseudonorms," I think. The corresponding unit sphere $\{ v : \langle Mv, v \rangle = 1 \}$ might now be lower-dimensional than the usual unit sphere, depending on how many eigenvalues are equal to zero; for example if $\dim V = 3$ it might be an ellipsoid, or an ellipse, or two points.

Let's consider the set $E$ of all vectors $y=Mx$, where $x\in\mathbb R^n$ belongs to the unit sphere (i.e. $\|x\|=1$). In other words, $E$ is the image of the unit sphere under the linear transformation. If the matrix is non-degenerate, $E$ is an $n$-dimensional ellipsoid. But if we assume additionally that $M$ is symmetric then we can say much more about the structure of $E$. Namely, the directions of the axes of the ellipsoid are pairwise orthogonal and represented by the eigenvectors of the matrix. Moreover, the lengths of the semiaxes are given by the corresponding eigenvalues.