Gram matrix invertible iff set of vectors linearly independent

Question

Given a set of vectors $v_1 \cdots v_n$, the $n\times n$ Gram matrix $G$ is defined as $G_{i,j}=v_i \cdot v_j$

Due to symmetry in the dot product, $G$ is Hermitian.

I'm trying to remember why $|G|=0$ iff the set of vectors are not linearly independent.

Answer 1

JasonMond's "only if" part is not as general as it should, because s/he assumes that $A$ is square. In the following, I complete the proof that holds whether $A$ is square or not.

Let $G = A^T A$. If the column vectors of $A$ are linearly dependent, there exists a vector $u \neq 0$ such that $$ A u = 0. $$ It follows that $$ 0 = A^T A u = G u. $$ Since $u \neq 0$, $G$ is not invertible.

Conversely, if $G$ is not invertible, there exists a vector $v \neq 0$ such that $$ G v = 0. $$ It follows that $$ 0 = v^T G v = v^T A^T A v = (A v)^T A v = \lVert A v\rVert^2 $$ and therefore that $$ A v = 0. $$ Since $v \neq 0$, the column vectors of $A$ are linearly dependent. QED.

Answer 2

Let $A$ be the matrix whose columns are the vectors $v_1, v_2, ... v_n$. Then the Gram matrix is $A^T A$, so $\det G = (\det A)^2$.

Edit: Here's an explanation that ignores the dimension of the ambient space. If $\langle \cdot, \cdot \rangle$ denotes the inner product, then the Gram matrix is precisely the matrix describing the inner product

$$\langle x_1 v_1 + ... + x_n v_n, y_1 v_1 + ... + y_n v_n \rangle$$

on $\mathbb{R}^n$. It's not hard to see that the vectors $v_i$ are linearly independent if and only if the above inner product is positive-definite. But we can write the above as $v^T Gv$ where $v \in \mathbb{R}^n$ and $G$ is the Gram matrix, and then we know that the inner product is positive-definite if and only if $G$ is invertible (since $G$ is invertible if and only if its eigenvalues are all positive).

Answer 3

Here's another way to look at it.

If $A$ is the matrix with columns $v_1,\ldots,v_n$, and the columns are not linearly independent, it means there exists some vector $u \in \mathbb{R}^n$ where $u \neq 0$ such that $A u = 0$. Since $G = A^T A$, this means $G u = A^T A u = A^T 0 = 0$ or that there exists a vector $u \neq 0$ such that $G u = 0$. So $G$ is not of full rank. This proves the "if" part.

The "only if" part -- i.e. if $|G| = 0$, the vectors are not linearly independent -- follows because $|G| = |A^T A| = |A|^2 = 0$ which implies that $|A| = 0$ and so $v_1,\ldots,v_n$ are not linearly independent.

Answer 4

For one thing, the Gram determinant is the square of the $n$-dimensional volume of the parallelopiped generated by the given basis vectors. The determinant is 0 when that object is not full dimensional, that is when one of the vectors lies in the hyperplane generated by the other $n-1$ vectors, in turn meaning linear dependence.

Answer 5

There is a simple proof that doesn't need the dimensionality of the ambient space.

Let $v_1,\ldots ,v_n$ a list of linearly independent vectors of an arbitrary Hilbert space, and set $W:=\operatorname{span}(v_1,\ldots ,v_n)$. Then $\dim W=n$ and so there is a Hilbert space isomorphism $T:W\to\mathbb C ^n$, just pick two orthonormal bases of $W$ and $\mathbb C ^n$ and map one in the other bijectively and extend the map linearly.

Let $A$ the $n\times n$ matrix who rows are $Tv_1\ldots ,Tv_n$, then its easy to check that $AA^\dagger$ (where $A^\dagger$ is the conjugate transpose of $A$) is the Gram matrix of $v_1,\ldots,v_n$. Therefore the $v_k$ are linearly independent if and only if $\det(A)\neq 0$, if and only if $\det(A^\dagger A)=(\det(A))^2\neq 0.\Box$

Answer 6

Let ${ H }$ be a real Hilbert space. Let ${ x _1, \ldots, x _n \in H . }$

Can we compute the space of linear dependencies amongst ${ x _1, \ldots, x _n }$?

Can we compute the space

$${ L = \lbrace \alpha \in \mathbb{R} ^n : \alpha _1 x _1 + \ldots + \alpha _n x _n = 0 \rbrace \, \, ? }$$

Let ${ \alpha \in L . }$ Note that taking inner products

$${ {\begin{aligned} &\, \alpha _1 (x _1 \, \vert \, x _1) + \ldots + \alpha _n (x _1 \, \vert \, x _n) = 0, \\ &\, \, \quad \quad \vdots \\ &\, \, \alpha _1 (x _n \, \vert \, x _1) + \ldots + \alpha _n (x _n \, \vert \, x _n) = 0 . \end{aligned}} }$$

Hence ${ \alpha \in \ker (G) }$ where ${ G = ((x _i \, \vert \, x _j)) _{i, j = 1} ^{n} . }$

Hence ${ L \subseteq \ker (G) . }$

Is ${ L = \ker (G) }$?

Let ${ \alpha \in \ker (G) . }$ Note that

$${ {\begin{aligned} &\, \alpha _1 (x _1 \, \vert \, x _1) + \ldots + \alpha _n (x _1 \, \vert \, x _n) = 0, \\ &\, \, \quad \quad \vdots \\ &\, \, \alpha _1 (x _n \, \vert \, x _1) + \ldots + \alpha _n (x _n \, \vert \, x _n) = 0 . \end{aligned}} }$$

Hence

$${ (\alpha _1 x _1 + \ldots + \alpha _n x _n \, \vert \, \alpha _1 x _1 + \ldots + \alpha _n x _n) = 0 . }$$

Hence

$${ \alpha _1 x _1 + \ldots + \alpha _n x _n = 0 .}$$

Hence ${ \alpha \in L . }$

Hence

$${ L = \ker(G) }$$

as needed.

Note that especially the space of linear dependencies amongst ${ x _1, \ldots, x _n }$ depends only on the inner products ${ ((x _i \, \vert \, x _j) ) _{i, j = 1} ^{n} . }$