# Invertibility Regarding Inner Product Spaces

Exercise Let $$(V, \langle \,\, , \,\,\rangle)$$ be an inner product space and let $$\mathcal{L} = \{v_1, \dots, v_n\} \subset V$$. Prove that $$\mathcal{L}$$ is linearly independent if and only if

$$A = \begin{bmatrix} \langle v_1, v_1 \rangle & \langle v_2, v_1 \rangle & \dots & \langle v_n, v_1 \rangle \\ \langle v_1, v_2 \rangle & \langle v_2, v_2 \rangle & \dots & \langle v_n, v_2 \rangle \\ \vdots & \vdots & \vdots & \vdots \\ \langle v_1, v_n \rangle & \langle v_2, v_n \rangle & \dots & \langle v_n, v_n \rangle \\ \end{bmatrix}$$

is invertible.

For the $$(\Rightarrow)$$ forward direction, if we suppose $$\mathcal{L}$$ is linearly independent, then we know that if

$$a_1v_1 + \dots a_nv_n = 0$$

then each $$a_i = 0, \hspace{0.4cm} 1 \leq i \leq n$$. Now, we know that orthogonal implies linear independent, but linear independence need not imply orthogonality. So it is tricky to see how we will arrive at the conclusion $$\det(A) \neq 0$$, i.e. $$A$$ is invertible.

For the $$(\Leftarrow)$$ backwards direction, we suppose that $$A$$ is invertible and so $$\det(A) \neq 0$$. Now, since $$\det(A) \neq 0$$, then there exists $$v_i, v_j \in \{v_1, \dots, v_n\}$$ so that

$$\langle v_i, v_j \rangle \cdot \det(A^*) \neq 0$$

where $$A^*$$ is the square matrix obtained from $$A$$ by eliminating the $$i^{th}$$ column and $$j^{th}$$ row. In other words, at least one term of the $$n \times n$$ determinant of $$A$$ is nonzero.

Are these approaches headed in the right direction or am I misled? I am unsure how to complete the proof for either direction. Any advice or suggestions are greatly appreciated in advance.

Hint. Let $$V=[v_1~\cdots~v_n]$$ be the matrix of column vectors $$v_i$$. Then the matrix in question is $$V^\dagger V$$ (assuming your complex inner products are conjugate-linear in the first argument - if we're talking about real inner product spaces we can just ignore the complex stuff). Notice $$a^\dagger(V^\dagger V)a=\|Va\|^2$$.
$$\begin{bmatrix} \overline{a_1} & \cdots & \overline{a_n} \end{bmatrix} \begin{bmatrix} \langle v_1,v_1\rangle & \cdots & \langle v_1,v_n\rangle \\ \vdots & \ddots & \vdots \\ \langle v_n,v_1\rangle & \cdots & \langle v_n,v_n\rangle \end{bmatrix} \begin{bmatrix} a_1 \\ \vdots \\ a_n \end{bmatrix} = \|a_1v_1+\cdots+a_nv_n \|^2$$
In fact, for real inner product spaces, $$\det(V^T V)$$ (the Grammian determinant) is the squared volume $$\mathrm{vol}^2$$ of the parallelotope spanned by $$v_1,\cdots,v_n$$. When the dimension of $$V$$ matches the number of vectors $$n$$, this is the special case that $$\det V=\mathrm{vol}$$, but otherwise it is more general.
This can be generalized even further to an inner product on the exterior power $$\Lambda V$$ which can be used to calculate the "volume distorion factor" associated with orthogonally projecting one subspace onto another.
• @EmilyBurkenhamen The hint reveals the hermitian matrix $V^\dagger V$ is positive-semidefinite, and moreover that $\{v_i\}$ is linearly independent iff $V^\dagger V$ is positive-definite. Keep in mind positive-definite matrices can be distinguished from the other positive-semidefinite matrices by their positive determinants. Dec 7 '21 at 16:09