Let (v_1,v_2,...,v_r) be a basis for an inner product space V. How can I show that the zero vector is the only vector in V that is orthogonal to all of the basis vectors?

Question

anonymous · Answer

Hmmm, well, any particular vector is going to be a linear combination of the basis vectors.

anonymous · Answer

A vector is not going to be orthogonal to a basis vector which it comes from.

anonymous · Answer

However, the basis are going to be orthogonal to each other.

anonymous · Answer

So lets consider any vector $\mathbf u$: $$
\mathbf u = \sum_{i=1}^rc_i\mathbf v_i
$$If $\mathbf u\neq \mathbf 0$, then that means there is some $c_i\neq 0$.

anonymous · Answer

We will suppose that $c_j\neq 0$.$$
\mathbf v_j \cdot \left(\sum_{i=1}^rc_i\mathbf v_i\right) = \mathbf v_j\cdot c_j\mathbf v_j
$$Because for all $i\neq j$ we get $\mathbf v_j\cdot c_i\mathbf v_i = 0$.

anonymous · Answer

Based on the properties of dot product, we can say: $$
\mathbf v_j \cdot \mathbf u = c_j\|\mathbf v_j\|^2\neq 0
$$