# Prove that Gramian matrix is Invertible iff $(v_1,…,v_k)$ is linearly independent

Prove that Gramian matrix is Invertible iff $$(v_1,...,v_k)$$ is linearly independent
$$G=G(v_1,...,v_k) = [\left\langle v_i,v_j\right\rangle ]_{i,j=1}^k$$
I have great idea to calculate $$\det G$$ and show that if $$(v_1,...,v_k)$$ is linearly independent then $$\det G \neq0$$ and vice versa.
Plan sounds good (?) but how to calculate $$\det$$ of this?
$$\begin{vmatrix} \langle v_1,v_1\rangle & \langle v_1,v_2\rangle &\dots & \langle v_1,v_n\rangle\\ \langle v_2,v_1\rangle & \langle v_2,v_2\rangle &\dots & \langle v_2,v_n\rangle\\ \vdots&\vdots&\ddots&\vdots\\ \langle v_n,v_1\rangle & \langle v_n,v_2\rangle &\dots & \langle v_n,v_n\rangle\end{vmatrix}$$. Probably I should use some scalar product propeties but it isn't clear for me how can I do that.

• Since $G=V^T V$, $\det(G)=\det(V)^2$ is non-zero iff $\det(V)$ is non-zero, i.e. iff the convex hull of $0,v_1,\ldots,v_n$ has a non-zero (oriented) volume. – Jack D'Aurizio Jan 2 '19 at 14:25

Let $$v_i=\sum_{k=1}^n a_{ki}e_k$$, where $$\{e_1,\dots,e_n\}$$ is an orthonormal basis. Then $$\langle v_i,v_j\rangle=\sum_{k}\sum_{l}a_{ki}a_{lj}\langle e_k,e_l\rangle= \sum_{k}a_{ki}a_{kj}$$ and therefore, if $$A=[a_{ij}]$$, we have $$G=A^TA$$.

If $$Gx=0$$, then also $$x^TA^TAx=(Ax)^T(Ax)=0$$, so $$Ax=0$$.

The matrix $$A$$ is the matrix (with respect to the basis $$\{e_1,\dots,e_n\}$$) of the linear map defined by $$e_i\mapsto v_i$$, for $$i=1,\dots,n$$. The rank of this matrix is $$n$$ if and only if $$\{v_1,\dots,v_n\}$$ is linearly independent.

If the set is linearly independent, then $$Ax=0$$ implies $$x=0$$; thus $$Gx=0$$ implies $$x=0$$ and $$G$$ has rank $$n$$.

If the set is not linearly independent, then there is $$x\ne0$$ with $$Ax=0$$, so also $$Gx=0$$ and $$G$$ is not invertible.

Notes. If the inner product is over the complex numbers, then $$G=A^HA$$ (the Hermitian transpose) and the argument goes through using the Hermitian transpose instead of the transpose. An orthonormal basis exists because of Gram-Schmidt algorithm.

• Thanks for clear solution – user617243 Jan 2 '19 at 18:39