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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 $v = \frac{\langle v,v_1\rangle}{||v_1||^2} v_1 + \cdots + \frac{\langle v,v_n\rangle}{\|v_n\|^2} v_n$.

I recognize this as the projection of $v_1$ onto $v +{}$ the projection of $v_2$ onto $v + \cdots +{}$ the projection of $v_n$ onto $v$ but I'm not sure how to show $v$ is equal to this. I'm assuming that because $β$ is a basis I can write any vector $v$ as a linear combination of its vectors (i.e. $v = c_1v_1 + \cdots + c_nv_n$), but then I don't know if I can show that $c_k = \frac{\langle v,v_k\rangle}{\|v_k\|^2}$ for every $k$. I think there's something about the basis being orthogonal that can help me reach my conclusion, but I can't seem to figure out exactly how to use this fact to help me.

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  • $\begingroup$ I'm the \langle \rangle fairy, here to let you know that $\langle, \rangle$ plays nicer with TeX than <, > does :) $\endgroup$ – Patrick Stevens Mar 23 '17 at 8:00
  • $\begingroup$ Thanks! Will use that in the future! $\endgroup$ – Chris Mar 23 '17 at 17:08
  • $\begingroup$ Note proper MathJax usage, as in my edits to this question. $\langle a, b\rangle$ is right; $<a,b>$ is not. Also $\|a\|$ rather than $||a||,$ $a+\cdots+z$ rather than $a+ ... + z,$ and $\dim(W)$ rather than $dim(W).\qquad$ $\endgroup$ – Michael Hardy Feb 3 '18 at 17:51
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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 $$ \langle v, v_i \rangle = \langle c_1v_1 + c_2v_2 + \cdots + c_nv_n, v_i\rangle = \langle c_iv_i, v_i\rangle = c_i \langle v_i, v_i \rangle = c_i\|v_i\|^2 $$

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    $\begingroup$ From $v = c_1v_1 + c_2v_2 + \cdots + c_nv_n$, taking inner product with $v_i$, we get $\langle v,v_i \rangle = c_i \langle v_i,v_i\rangle = c_i\|v_i\|^2$, since $\langle v_i , v_j \rangle = 0$ when $i \neq j$ $\endgroup$ – user348749 Mar 23 '17 at 7:49
  • $\begingroup$ @Muralidharan Thanks. It makes it more clear. $\endgroup$ – PSPACEhard Mar 23 '17 at 7:53

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