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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

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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.

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  • $\begingroup$ Any comment about the correctness of this proof is welcome $\endgroup$ – hardhu Feb 4 '16 at 13:24
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Write ${\bf V}=({\bf v_1},...,{\bf v_m})$ then $\bf P=V^TAV$. Then using \begin{equation} 0\neq\prod_i {\bf P}_{ii} = det({\bf P})=det({\bf V^T})det({\bf A})det({\bf V})=det({\bf V})^2det({\bf A}). \end{equation} And therefore $det({\bf V})\neq0$. In other words the vectors ${\bf v_1},...,\bf{v_m}$ are linearly independent. Your second equation implies that ${\bf A}$ is non-singular, otherwise $0\neq\prod_i {\bf P}_{ii}$ would conflict with $det {\bf A}=0$.

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