Orthonormal set and linear independence

Question

My book has this Theorem:

Let $A=\{v_1,v_2,\dots,v_n\}$ be an orthonormal subset of an inner product space $X$. Prove that $A$ is also linearly independent.

The proof it provides starts like this : $$\sum_{k=1}^n λ_kv_k=0\implies \left\langle\,\sum_{k=1}^n λ_kv_k,v_1\,\right\rangle=0$$ and then it shows that $λ_1=0$ and similarly you prove the same for the rest of $λ_k$. Can you explain to me how this implication works? How you go from $\sum_{k=1}^n λ_kv_k=0$ to $\left\langle\,\sum_{k=1}^n λ_kv_k,v_1\,\right\rangle=0$ ?

Thank you very much in advance.

Answer 1

Since $\sum_{k=1}^n\lambda_kv_k=0$,$$\left\langle\sum_{k=1}^n\lambda_kv_k,v_1\right\rangle=\langle0,v_1\rangle=0.$$

Answer 2

Use directly the aximos of the inner product.

You use that $<\sum u_i,w>=\sum<u_i,w>$. Since your set is orthonormal $<u_i,u_j>=δ^i_j$.

Can you continue from here ?

Answer 3

As was said in the comments below your question, the essential assumption you did not included is that the set $A$ is composed of orthonormal vectors, meaning $<v_i, v_j> = 0$ for $ i \neq j$ and $<v_i, v_i> \neq 0$ for all $i \in \{1, \dots, n \} $. Then the proof goes as follows.

Since the inner product $< \cdot , \cdot > $ is bilinear (i.e. linear in each of its two slots) we have $$ < \sum_{k = 1}^n \lambda_k v_k, v_i> = \sum_{k = 1}^n \lambda_k < v_k, v_i> \ .$$ Using the orthonormality of the set $A$ we know that the above is equal to $$ \sum_{k = 1}^n \lambda_k < v_k, v_i> = \lambda_i < v_i, v_i> \ .$$ From the assumption $< \sum_{k = 1}^n \lambda_k v_k, v_i> = 0$ we get $$ \lambda_i < v_i, v_i> = 0 $$ and thus $\lambda_i$ must be zero since $< v_i, v_i> \neq 0$. Repeating this procedure with different $v_k$ in the right slot of $< \cdot, \cdot > $ gives the proof of the statement.