Let $V$ be an Inner Product Space with basis $v_1,v_2, \cdots , v_n$. Prove that if $<x,v_k>=0$ for all $k$, then $x=0$

Question

Since $v_1,v_2, \cdots , v_n$ forms a basis of $V$, I know that I can write $x=\lambda_1v_1+\cdots+\lambda_nv_n$. So the inner product is now $<\lambda_1v_1+\cdots+\lambda_nv_n, v_k>$. By the linearity of the inner product, this can be expanded, but I'm not sure where to go from there.

Answer 1

How about: $<x,x>= \sum_k <x,\lambda_k v_k> = \sum_k \lambda_k <x,v_k> = 0$

Answer 2

Hint: you can argue by contradiction: suppose x would be nonzero. Then you can show that the set $v_1, \dots v_n, x$ (i.e. the basis enlarged by the vector $x$) is linearly independent and thats a contradiction (because the maximal number of linearly independent vectors is $n$).

Answer 3

We can use this property of the inner product:

• If $\langle x,v \rangle = 0$ for all $v \in V$, then $x = 0$.

To show that the above condition holds, given any $v \in V$ you can expand it out using the basis. Then by linearity, $\langle x,v \rangle$ is a linear combination of $\langle x,v_1 \rangle, \ldots, \langle x,v_n \rangle$, which must be zero.