TITLE: Linear independence of matrix-conjugate vectors
QUESTION [5 upvotes]: Let's define the vectors $\mathbf{v}_1,\dots,\mathbf{v}_m \in \mathbb{R}^n$, with $m\leq n$, to be mutually conjugate with respect to matrix $\mathbf{A} \in \mathbb{R}^{n\times n}$ if 
$$\mathbf{v}_{i}^T\mathbf{A}\mathbf{v}_j = 0, \qquad 1 \leq i,j \leq m, \qquad i\not=j.$$ Assume that $\mathbf{A}$ is nonsingular, and suppose that 
$$ \mathbf{v}_{i}^T\mathbf{A}\mathbf{v}_i \not= 0, \qquad 1 \leq i\leq m.$$
Prove then that $\mathbf{v}_1,\dots,\mathbf{v}_m$ are linearly independent in $\mathbb{R}^n$. Does this hold even if $\mathbf{A}$ is singular?
Idea of the proof
The sketch of the proof I have found suggest to consider the matrix 
$$\mathbf{P}=\begin{pmatrix}
 \mathbf{v}_1 \dots \mathbf{v}_m \end{pmatrix}^T \mathbf{A} \begin{pmatrix}
\mathbf{v}_1 \dots \mathbf{v}_m \end{pmatrix}.$$
It is a diagonal matrix, with $$\mathbf{P}_{(i,i)}=\mathbf{v}_{i}^T\mathbf{A}\mathbf{v}_i$$ then its determinant is not null. How could I infer then that the vectors are linearly independent? Is it necessary the assumption $\mathbf{A}$ nonsingular

REPLY [2 votes]: Ok, I seem to have found a different proof that doesn't require $\mathbf{A}$ to be nonsingular. Indeed, suppose that $\mathbf{v}_1,\dots,\mathbf{v}_m$ are not linearly independent. Then there exist $k \in \{1,\dots,m\}$ such that 
$$ \mathbf{v}_k=\sum_{\substack{i=1\\i\not=k}}^m\alpha_i\mathbf{v}_i$$ and for at least $h\in \{1,\dots,m\}, h\not=k$ it is $\alpha_h \not=0$. Then we have
$$ \mathbf{v}_{h}^T\mathbf{A}\mathbf{v}_k=\mathbf{v}_{h}^T\mathbf{A}\left(\sum_{\substack{i=1\\i\not=k}}^m\alpha_i\mathbf{v}_i\right)=\sum_{\substack{i=1\\i\not=k}}^m \alpha_i\mathbf{v}_{h}^T\mathbf{A}\mathbf{v}_i=\alpha_h\mathbf{v}_{h}^T\mathbf{A}\mathbf{v}_h\not=0$$
a contradiction.