Verify the pythagorean theorem for three vectors.

Question

Let $V \in \mathbb{R}^3$ and let $\langle u,v \rangle = u^TAv$ where $A = \begin{bmatrix}2&0&1\\0&2&0\\1&0&2 \end{bmatrix}$

Let $f_1 = (1,0,-1)^T, f_2 = (0,1,0)^T, f_3 = (1,0,1)^T$ be an orthogonal basis for $V$.

Verify the Pythagorean Theorem for the sum $f_1 + f_2 + f_3$

My Attempt:

My initial attempt was to compare the norm of $f_1 + f_2 + f_3 = f_4$ that is, $(\sqrt{\langle f_4, f_4\rangle})$ against its eucledian distance from the zero vector, but I believe this is completely incorrect. I am not sure how to proceed with this.

Apologies for not providing too much attempts as I am not sure about the other ways I could approach this.

Answer 1

Hint: Following your notation of $f_4 = f_1 + f_2 + f_3$, we are meant to check that $$ \langle f_1,f_1 \rangle + \langle f_2,f_2 \rangle + \langle f_3,f_3 \rangle = \langle f_4,f_4 \rangle $$ Euclidean distance has no relevance here. However, note that $\langle u,u\rangle$ is the square of the distance induced by this new inner product.