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Let V be an inner product space and $S={v_1,v_2,...v_k}$ be an orthogonal subset of V consisting of nonzero vectors. If $y \in \operatorname{span}(S)$, then $y=\sum_{i=1}^k \frac{\langle y,v_i\rangle}{||v_i||^2} v_i$.

For $y=\sum_{i=1}^k \frac{\langle y,v_i\rangle}{||v_i||^2} v_i$, can someone explain the relation between $\frac{\langle y,v_i\rangle}{||v_i||^2}$ and $\cos \theta =\frac{\langle a,b\rangle}{|a|\cdot |b|}$ ? So does that mean $\frac{\langle y,v_i\rangle}{||v_i||^2}$ is length of orthogonal vector projected onto the subspace A ?enter image description here

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    $\begingroup$ Which plane are you talking about at the end of your question? $\endgroup$ Mar 11 '20 at 16:36
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    $\begingroup$ Use \langle and \rangle instead of < and > for inner product brackets. $\endgroup$ Mar 11 '20 at 16:37
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The length of the projection of $y$ onto the one-dimensional subspace spanned by the single vector $v_i$ is $\|y\| \cos \theta$ where $\theta$ is the angle between $y$ and $v_i$. (This is just an observation/definition from trigonometry.)

You know that $\cos \theta = \frac{\langle y, v_i \rangle}{\|y\| \|v_i\|}$, so the length of the vector is $\frac{\langle y, v_i \rangle}{\|v_i\|}$. However, the actual vector of the projection is this length times the unit vector in that direction, so it is $\frac{\langle y, v_i \rangle}{\|v_i\|} \frac{v_i}{\|v_i\|} = \frac{\langle y, v_i \rangle}{\|v_i\|^2} v_i$.


The formula for projecting onto $\text{span}\{v_1, \ldots, v_k\}$ will in general require some additional information about the $v_i$. But if you know that they are orthogonal, it turns out that the projection is the sum of each of the projections onto the one-dimensional subspaces $\text{span}\{v_i\}$.

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