# About members of a basis of an inner product space

Say we have an inner product space $V$ with a basis $\mathbb B=\{v_1,...,v_n\}$, can we say that for each $i,j\in[1,n],i\neq j$ we have $\langle v_i,v_j \rangle=0$ because it's an inner product space?

$\langle x,y\rangle$ is the inner product.

• No.${}{}{}{}{}$ – David Mitra Nov 9 '15 at 17:00
• What can we say about the vectors in the basis then? Can we assume anything about them? @DavidMitra – shinzou Nov 9 '15 at 17:01
• They are independent. Not much else ... – David Mitra Nov 9 '15 at 17:02
• A basis does not have to be an orthogonal basis just because it is an innner product space. – Nyfiken Nov 9 '15 at 17:07

Take $V=\mathbb R^2$ and $v_1=(1,0)$ and $v_2=(1,1)$. Then $\{ v_1, v_2 \}$ is a basis but $\langle v_1, v_2 \rangle = 1 \ne 0$.
One thing that you can say about a generic basis is that its Gramian is not zero: $$\begin{vmatrix} \langle v_1,v_1\rangle & \langle v_1,v_2\rangle &\dots & \langle v_1,v_n\rangle\\ \langle v_2,v_1\rangle & \langle v_2,v_2\rangle &\dots & \langle v_2,v_n\rangle\\ \vdots&\vdots&\ddots&\vdots\\ \langle v_n,v_1\rangle & \langle v_n,v_2\rangle &\dots & \langle v_n,v_n\rangle\end{vmatrix} \ne 0$$