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Let $\langle x,y\rangle$ be an inner product on $F^m$ and define $\|x\| = \sqrt{\langle x,x\rangle}$. Suppose $W$ is a subspace of $F^m$ and $\dim(W) = n$.

Let $β = \{v_1, v_2, \ldots , v_n\}$ be an orthogonal basis for $W$. Prove that for any vector $v ∈ W$ we have $v = \frac{\langle v,v_1\rangle}{||v_1||^2} v_1 + \cdots + \frac{\langle v,v_n\rangle}{\|v_n\|^2} v_n$.

Proof: Denoting $v = c_1v_1 + c_2v_2 + \cdots + c_nv_n$ for some $c_1, c_2, \cdots, c_n$ since $\{v_1, v_2, \cdots, v_n\}$ is a basis. We have \begin{align} & v - \frac{\langle v, v_1\rangle}{\|v_1\|^2}v_1 - \cdots - \frac{\langle v, v_n\rangle}{\|v_n\|^2}v_n \\ =\ &\left(c_1 - \frac{\langle v, v_1\rangle}{\|v_1\|^2}\right)v_1 + \cdots + \left(c_n - \frac{\langle v, v_n\rangle}{\|v_n\|^2}\right)v_n \\ =\ & \left(c_1 - \frac{\langle c_1v_1, v_1\rangle}{\|v_1\|^2}\right)v_1 + \cdots + \left(c_n - \frac{\langle c_nv_n, v_n\rangle}{\|v_n\|^2}\right)v_n \\ =\ &\left(c_1 - c_1\right)v_1 + \cdots + \left(c_n - c_n\right)v_n \\ =\ &0 \end{align} where the second equality is because $$ \langle v, v_i \rangle = \langle c_1v_1 + c_2v_2 + \cdots + c_nv_n, v_i\rangle = \langle c_iv_i, v_i\rangle = c_i \langle v_i, v_i \rangle = c_i\|v_i\|^2 $$

I am not sure how it arrives from $\left(c_1 - \frac{\langle c_1v_1, v_1\rangle}{\|v_1\|^2}\right)v_1 + \cdots + \left(c_n - \frac{\langle c_nv_n, v_n\rangle}{\|v_n\|^2}\right)v_n$ to $\left(c_1 - c_1\right)v_1 + \cdots + \left(c_n - c_n\right)v_n$

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1 Answer 1

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$\frac{\langle c_1 v_1, v_1 \rangle}{\Vert v_1 \Vert^2} = \frac{c_1\langle v_1, v_1 \rangle}{\Vert v_1 \Vert^2} = c_1$ since $\langle v_1, v_1 \rangle = \Vert v_1 \Vert^2$ per definition. Analogous for $v_2,...,v_n$.

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