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Let $V$ be a real inner product space with basis $U=(v_1, \ldots, v_n)$ and $A \in Mat_{n,n}(\mathbb R)$ with $(i,j)$-entry equal to $\langle v_i v_j \rangle$.

I've shown the identity $\langle x, y \rangle = [x]_U^T \cdot A \cdot [y]_U \ \forall x,y \in V$ by writing $x= c_1 v_1 + \ldots c_n x_n$ and $y= d_1 v_1 + \ldots d_n v_n$, and then expanding inner product using axioms and afterwards computing the matrix product.

However I must show $A$ is invertible.

I try to show this by writing $Ax = 0$ and show $x = 0$.

$Ax = 0 \iff x_1 \langle v_i, v_1 \rangle + \ldots + x_j \langle v_i, v_j \rangle + \ldots \langle v_i, v_n \rangle = 0 \iff \langle v_i, x_1 v_1 + \ldots + x_j v_j + \ldots + x_n v_n \rangle = 0$

I want to conclude $x_1 v_1 + \ldots + x_j v_j + \ldots + x_n v_n = 0$ and then since $v_i : 1 \le i \le n$ is a basis we have $x_j = 0 : 1 \le j \le n$. However, how can I do this ?

By the way, the matrix $A$ is called the Gram matrix for $v_1, \ldots, v_n$.

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  • $\begingroup$ Do you know that you can orthogonalize bases in finite dimensional real inner product spaces? $\endgroup$ – k.stm Mar 10 '14 at 9:37
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What happens is that your vector $v=a_1x_1+\ldots +a_nx_n$ is orthogonal to all the $v_i$. So $v$ is orthogonal to any linear combination of the $v_i$, so $v$ is orthogonal to itself. Since $<,>$ is an inner product, this is only possible if $v=0$.

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  • $\begingroup$ Depends on whether $V = ℝ^n$ and “$〈–,–〉$” denotes the euclidean inner product. $\endgroup$ – k.stm Mar 10 '14 at 9:44
  • $\begingroup$ I think $[y]_U$ should denote the coordinates of $y$ in base $U$. This identity holds for abstract inner product spaces as well. $\endgroup$ – k.stm Mar 10 '14 at 9:48
  • $\begingroup$ @k.stm please see my revised version. $\endgroup$ – Ewan Delanoy Mar 10 '14 at 9:54
  • $\begingroup$ Ewan why is it only possible if $v = 0$ that $v$ is ortogonal to every $v_i$ ? $\endgroup$ – Shuzheng Mar 10 '14 at 9:55
  • $\begingroup$ Maybe this is yet to prove! $\endgroup$ – k.stm Mar 10 '14 at 9:57
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There is a base change $S$ such that $US$ is a orthogonal base. You can prove this by induction on the dimension $n$. Let $v_1', …, v_n'$ be the transformed orthogonal base $US$. Calculate that $S^TAS = (〈v_i',v_j'〉)_{i,j = 1, …,n}$. To do this, you can use the identity you already have proven.

Here $US$ denotes $(s_{1,1}v_1 + …, s_{n,1}v_n, …, s_{1,n}v_1 + … s_{n,n}v_n)$, i.e. the system obtained by using the columns of $S$ to linearly combine out of $(v_1, …, v_n)$.

Because $A$ is invertible if and only if $S^TAS$ is invertible, you can assume without loss of generality your base to be an orthogonal system. Then your argument will work.

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Let $(a_1,...,a_n)$ be the vector which makes it 0. Then in the first row $v_1(a_1 v_1+.... a_n v_n)=0$ implies $a_1=0$. Similarly you can show all $a_n$ are 0.

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  • $\begingroup$ Depends on whether $V = ℝ^n$ and “〈–,–〉” denotes the euclidean inner product. $\endgroup$ – k.stm Mar 10 '14 at 9:47

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