Definiteness of Gram matrix
Definition 1 Definiteness of a matrix
Let $A$ be $n \times n$ real symmetric matrix, and column vector $x \in \mathbb{R}^n$.
- $A$ is positive (negative) definite if $x^TAx > 0 \ (x^TAx < 0),\ \forall x \neq 0$.
- $A$ is positive (negative) semidefinite if $x^TAx \geq 0 \ (x^TAx \leq 0),\ \forall x \in \mathbb{R}^n $, and $x^TAx = 0$ for at least one $ x \neq 0$.
- $A$ is indefinite if $x^TAx$ takes on both positive and negative values (with different, non-zero $x$'s)
Definition 2 Gram matrix
The Gram matrix of a set of vectors $v_1, \dots, v_n$ in an inner product space is the Hermitan matrix of inner products, whose entries are given by (1).
$$
\begin{align}
G_{ij} = \langle v_i, v_j \rangle \tag{1} \\
\end{align}
$$
For finite dimensional real vectors in $\mathbb{R}^n$ with the usual Euclidean dot product, the Gram matrix $G$ is simply $G=V^TV$, where the column of V is $v_i$.
$$
\begin{align}
G_{ij} = \langle v_i, v_j \rangle \tag{1} \\
\end{align}
$$
For finite dimensional real vectors in $\mathbb{R}^n$ with the usual Euclidean dot product, the Gram matrix $G$ is simply $G=V^TV$, where the column of V is $v_i$.
Theorem 1 The symmetric matrix $G$ is a Gram matrix if and only if it is positive semidefinite or positive definite.
Proof
(1) Gram matrix $G=V^TV$ is positive semidefinite or positive definite.
$$x^TAx = x^TV^TVx = (Vx)^T(Vx) = \| Vx\|^2 \geq 0, \forall x \in \mathbb{R}^n
$$
(2) Positive semidefinite or positive definite matrix is Gram matrix.
Positive semidefinite or positive definite $n \times n$ matrix $G$ with rank $r \leq n$ can be decomposed as (2)-(a) by Spectral Theorem (since those are symmetric, hence Hermitan), where $T = diag(\lambda_1, \dots , \lambda_n)$ is the diagonal matrix containing the eigenvalues in its main diagonal, while $U = (u_1, \dots , u_n)$ is the orthogonal matrix containing the corresponding eigenvectors of $A$ in its columns in the order of the eigenvalues.
$$
G \stackrel{(a)}{=} UTU^H = U \sqrt{T} \sqrt{T} U^H = U \sqrt{T}^H (U \sqrt{T}^H)^H \tag{2} \\
$$
Theorem 2 Gram matrix $G=V^TV$ is positive definite if and only if the column vectors in $V$ are linearly independent (or ker V = {0}).
Reference
Horn, Roger A.; Johnson, Charles R. (2013). "7.2 Characterizations and Properties". Matrix Analysis (Second ed.). Cambridge University Press. ISBN 978-0-521-83940-2.
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