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 $a^\dagger(V^\dagger V)a=\|Va\|^2$.

In more explicit terms (and also coordinate-free if you don't want to write vectors as columns):

$$ \begin{bmatrix} \overline{a_1} & \cdots & \overline{a_n} \end{bmatrix} \begin{bmatrix} \langle v_1,v_1\rangle & \cdots & \langle v_1,v_n\rangle \\ \vdots & \ddots & \vdots \\ \langle v_n,v_1\rangle & \cdots & \langle v_n,v_n\rangle \end{bmatrix} \begin{bmatrix} a_1 \\ \vdots \\ a_n \end{bmatrix} = \|a_1v_1+\cdots+a_nv_n \|^2 $$

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.

  • $\begingroup$ What does this have to do with the proposed problem? It's not obvious what this is asserting $\endgroup$ Dec 7 '21 at 13:37
  • $\begingroup$ @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. $\endgroup$
    – runway44
    Dec 7 '21 at 16:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.