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Let $V$ be an inner product space with $\mathcal{O} = \{u_1, \ldots, u_n\}$ serving as an orthonormal subset of $V$. Let $x$ be a vector in $V$ and let $\hat{x} = \left\langle u_1, x \right\rangle u_1 + \ldots + \left\langle u_n, x \right\rangle u_n$. Why is it the case that $\left\langle x, u_k \right\rangle = \left\langle \hat{x}, u_k \right\rangle$?

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  • $\begingroup$ Perhaps there is some subtlety here, but why isn't it the case that $\hat x = x$? It looks to me like it's just breaking down $x$ into components on a basis. $\endgroup$ – Muphrid May 9 '13 at 4:58
  • $\begingroup$ @Muphrid, I think it's because $\mathcal{O} = \{u_{1}, \ldots, u_{n}\}$ is given to be an orthonormal subset, not a basis. The result holds true even if $\mathcal{O}$ is NOT a basis, whereas $\hat{x} = x$ only if $\mathcal{O}$ is a basis. $\endgroup$ – Alex Wertheim May 9 '13 at 5:12
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The answer is that your inner product $\langle \cdot,\cdot\rangle$ is linear in each coordinate. Take $\langle \hat{x},u_k\rangle$, replace $\hat{x}$ by the expression, and expand using the linear property. All the terms will be zero except $\langle u_k,u_k\rangle$, with a coefficient of $\langle u_k,x\rangle$.

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Hint: what is $\langle u_{i}, u_{k} \rangle$ for any $i \neq k$ in an orthonormal subset? What is $\langle u_{k}, u_{k} \rangle$?

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