# Gram matrix invertible iff set of vectors linearly independent

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.

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.

• Beautiful proof! – Johnny Ji Nov 28 '17 at 9:26

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).

• This only works if $A$ is square which assumes that there are as many vectors in the set as there are dimensions in each vector. Consider for example $v_1=<1,0,1>$ and $v_1=<0,1,1>$. Then $G=\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array}$, whose determinant is non-zero. – Gus May 3 '11 at 4:21
• @Gus: if $v_1, ... v_n$ live in an inner product space of dimension less than $n$, then they are never linearly independent. If they live in an inner product space of dimension greater than $n$, we can find an orthonormal basis of the subspace that they span, so we can assume that they live in an $n$-dimensional inner product space without loss of generality. – Qiaochu Yuan May 3 '11 at 4:24
• @Qiochu I know, by Gram-Schmidt, how to find the orthonormal basis that you mentioned. I can then represent each vector in my original set by this basis, thereby reducing the dimension of the vectors to equal the dimension of the subspace that they span. Do you think there is a way to demonstrate the property I'm asking about without requiring this reduction? – Gus May 3 '11 at 4:33
• @Gus: sure. I'll edit in another explanation. – Qiaochu Yuan May 3 '11 at 4:39
• I found this looking for help with a proof, I want to show det(G)=det(A)$^2$ essentially. but I cannot assume that my linearly independent set of vectors form a basis for the space they come from, they might but the could be the basis for a subspace too. However as n doesn't matter (it must be greater than or equal to the number of linearly independent vectors I have by definition) I believe I can assume n=number of linearly indepdent vectors - so A^T and A are square, so det makes sense, right? Because that n works to make the product be the square matrix G – Alec Teal Nov 25 '13 at 0:42

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.

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.

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 $$\det(A^\dagger A)=(\det(A))^2\neq 0.\Box$$