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Let $(V,\langle \cdot,\cdot\rangle)$ be an inner product space over the real field $\mathbb{R}$ and $v_1,\dots,v_n\in V$.
Suppose that $A=(a_{ij})\in M_n(\mathbb{R)}$ with $a_{ij}=\langle v_i,v_j\rangle$ for all $1\leq i,j\leq n$.
Show that if $\det A=0$, then $v_1,\dots,v_n$ are linearly dependent in $V$.

Here I try to work out for $M_2(\mathbb{R})$. So $$A=\begin{pmatrix}\langle v_1,v_1\rangle &\langle v_1,v_2\rangle\\\langle v_1,v_2\rangle &\langle v_i,v_j\rangle\end{pmatrix}$$ Since its determinant is zero, we get $$||v_1||^2||v_2||^2-\langle v_1,v_2\rangle^2=0$$ So here I don't have idea how to show that $v_1=\alpha v_2$ for some $\alpha\in \mathbb{R}$.

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A basic result from linear algebra tells us that because the determinant of $A$ is $0$, the vectors in $\mathbb{R}^n$ that comprise the columns of $A$ must be linearly dependent. This means that: \begin{eqnarray*} \lambda_1\begin{pmatrix}\langle v_1,v_1\rangle\\ \langle v_1,v_2\rangle\\\vdots\\ \langle v_1,v_n\rangle \end{pmatrix}+ \lambda_2\begin{pmatrix}\langle v_2,v_1\rangle\\ \langle v_2,v_2\rangle\\\vdots\\ \langle v_2,v_n\rangle \end{pmatrix}+ \cdots+ \lambda_n\begin{pmatrix}\langle v_n,v_1\rangle\\ \langle v_n,v_2\rangle\\\vdots\\ \langle v_n,v_n\rangle \end{pmatrix}=\begin{pmatrix}0\\0\\\vdots\\0\end{pmatrix} \end{eqnarray*} so that for each $i$, $\lambda_1\langle v_1,v_i\rangle+\ldots+\lambda_n\langle v_n,v_i\rangle=0$. Using the properties of the inner product we get that $\langle \lambda_1v_1+\ldots+\lambda_nv_n,v_i\rangle=0$ for all $i$. This implies that $\lambda_1v_1+\ldots+\lambda_nv_n=0$ as desired.

Why does it follows that $\lambda_1v_1+\ldots+\lambda_nv_n=0$? The reason is because the inner product of $\lambda_1v_1+\ldots+\lambda_nv_n$ with all of its "components" is $0$. More explicitly, because $\langle \lambda_1v_1+\ldots+\lambda_nv_n, v_i\rangle=0$ it follows that $\langle \lambda_1v_1+\ldots+\lambda_nv_n, \lambda_iv_i\rangle=0$. Thus, adding all of these inner product expressions, we find that: \begin{eqnarray*} \langle \lambda_1v_1+\ldots+\lambda_nv_n, \lambda_1v_1+\ldots+\lambda_nv_n\rangle=0 \end{eqnarray*} which is only possible if $\lambda_1v_1+\ldots+\lambda_nv_n=0$.

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  • $\begingroup$ Nice. This answer is clear and simple. $\endgroup$ – Wes Jun 18 '16 at 3:31

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