pythagorean theorem Calculus

Question

Let be $\hat x_1,......,\hat x_k \in R^n$ \ $\hat {0}$ such that $ \hat{x_i} \cdot \hat {x_j} =0$ if $i,j \in$ {$1,....,k$},$ i\neq j$. Prove that

$$\Vert \hat x_1....+\hat x_k \Vert^2 = \Vert \hat x_1 \Vert^2+.....+\Vert \hat x_k \Vert^2$$.

I started using triangle inequalty and then proving that the strict inequalty does not stand but I could not do it. Some help would be just nice.

Answer 1

HINT

By dot product $\|v\|^2=v\cdot v$ we have

$$\Vert \hat x_1+...+\hat x_k \Vert^2 = (x_1+...+\hat x_k)\cdot(x_1+...+\hat x_k)$$

Answer 2

Just write that :

$$ \left \| \sum_{i=1}^k \hat{x_i} \right \|^2 = \left\langle \sum_{i=1}^k \hat{x_i},\sum_{i=1}^k \hat{x_i} \right\rangle$$

When developped it leads to the result, you just have to use the bilinearity of $\langle . , . \rangle$.