# Using Inner Product: If $X$ is nonsingular, then $Y$ is linearly independent over $F$

Problem:

Let $F = \mathbb{R} \textrm{ or } \mathbb{C}$ and let $_FV$ be finite dimensional with inner product $\langle\cdot{,}\cdot\rangle$. Suppose $k$ is a positive integer and $Y = \{v_1,v_2,\ldots, v_k\} \subset V$. Let $X = [x_{ij}] \in M_k(F)$ such that $x_{ij}$ = $\langle v_i {,} v_j \rangle$, for all $i$ and $j$.

To show: If $X$ is nonsingular, then $Y$ is linearly independent over $F$.

Can anyone provide a hint for the proof? I've tried to prove by contradiction. I've used the assumption being nonsingular so that should have linearly independent columns and then considered orthogonality. Is my approach wrong?

Let $n=\hbox{dim}V$. The assertion is clearly false if $k>n$, so let's assume $k\leq n$, and furthermore, assume first that $k=n$. Write each $v_k$ by it's coordinates (either real or complex): $v_k=(v_k(1),\dots,v_k(n))$ and let $A$ denote the $n\times n$ matrix whose rows are the vectors $v_1,v_2,\dots v_n$. That is: $$A=\begin{pmatrix}v_1(1)&v_1(2)&\dots &v_1(n)\\ v_2(1)&v_2(2)&\dots&v_2(n)\\ \vdots&\vdots&\dots&\vdots\\ v_n(1)&v_n(2)&\dots &v_n(n)\end{pmatrix}$$

Observe that $AA^t$ is the matrix whose $(i,j)$ entry is precisely the inner product $\langle v_i,v_j\rangle$. Since $\det AA^t=(\det A)^2$, we see that $A$ is non-singular if and only if $AA^t$ is non-singular. But $A$ is non-singular if and only if the vectors $v_1,\dots v_n$ are linearly independent. This proves the assertion for the case $k=n$.

If $k<n$, we would like to add vectors to $v_1,\dots v_k$, say, $u_1,\dots u_{n-k}$, such that all the $u_i$'s are orthogonal to all the $v_i$'s, and the $u_i$'s are normalised and mutually orthogonal. We can find such $n-k$ vectors because the dimension of the span of the $v_i$'s is at most $k$, so that the dimension of it's orthogonal complement is at least $n-k$. Now our matrix will look like this:

$$A=\begin{pmatrix}v_1(1)&v_1(2)&\dots &v_1(n)\\ \vdots&\vdots&\dots&\vdots\\ v_k(1)&v_k(2)&\dots&v_k(n)\\ u_1(1)&u_1(2)&\dots &u_1(n)\\ \vdots&\vdots&\dots&\vdots\\ u_{n-k}(1)&u_{n-k}(2)&\dots &u_{n-k}(n) \end{pmatrix}$$ and when we will multiply $A$ by it's transpose $A^t$ we will get this matrix:

$$AA^t=\begin{pmatrix}\langle v_1,v_1\rangle &\dots &\langle v_1,v_k\rangle& 0&\dots & 0\\ \vdots&\vdots&\dots&\vdots&\ddots\\ \langle v_k,v_1\rangle &\dots &\langle v_k,v_k\rangle& 0&\dots & 0\\ 0&\dots &0&1&\dots&0\\ \vdots&\ddots&\dots&\vdots&\ddots&\vdots\\ 0&\dots&0 &0&\dots & 1 \end{pmatrix}$$

so the block on the bottom right side is the $(n-k)\times (n-k)$ identity matrix. Now, since the $u_i$'s are linearly independent and orthogonal to the $v_i$'s, we see that the $v_i$'s are linearly independent if and only if $\det A\neq 0$, and this happens if and only $\det AA^t \neq 0$, which is equal to the determinant of the $k\times k$ matrix $\langle v_i,v_j\rangle$. This completes the proof for the case $k<n$ also.

It is not strictly necessary to separately prove the case $k=n$, but it seems to illustrate well what's going on.