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Given a vector space V where two vectors v and u are orthogonal with respect to a given inner product. Will the same vectors be orthogonal with respect to all the other inner products?

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In general, no. Take $\mathbb{R}^2$ with its usual inner product and consider also the inner product$$\bigl\langle(x_1,x_2),(y_1,y_2)\bigr\rangle=2x_1y_1-x_1y_2-x_2y_1+x_2y_2.$$Then $(1,0)$ and $(0,1)$ are orthogonal with respect to the usual inner product, but $\bigl\langle(1,0),(0,1)\bigr\rangle=-1$.

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In fact, the orthogonality relation specifies an inner product up to a positive scalar multiple. To see this, suppose $\langle \cdot, \cdot \rangle$ and $[ \cdot, \cdot ]$ are two inner products on a space $V$ with the same orthogonality relation. Note that if $V = \lbrace 0 \rbrace$, then the two inner products must be the same, so otherwise $V$ contains non-zero vectors.

Let $w \in V$ be non-zero, and define $k_w = \frac{\langle w, w \rangle}{[w, w]} \in (0, \infty)$. Clearly, if $v = \lambda w$ for scalar $\lambda$, then $\langle v, w \rangle = k_w [v, w]$. Otherwise, if $v$ is independent from $w$, then $$v - \frac{[v, w]}{[w, w]}w \perp w$$ under both inner products, in particular, $$0 = \left\langle v - \frac{[v, w]}{[w, w]}w, w \right\rangle \implies \langle v, w \rangle = \frac{\langle w, w \rangle}{[w, w]} [v, w] = k_w [v, w].$$ By a corresponding argument, we can also formulate $k_v \in (0, \infty)$ instead, and show $$\langle w, v \rangle = k_v [w, v],$$ which implies that $k_v = k_w$. This holds for linearly independent $v, w$ as well as linearly dependent $v, w$ as shown above. So $k_v$ and $k_w$ don't depend on $v$ and $w$ at all, and is in fact a constant. Thus, some $k \in (0, \infty)$ exists such that $$\langle w, v \rangle = k [w, v],$$ as required.

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