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I'm given the inner product $f[(x,y,z),(a,b,c)]=2yb+yc+zb+xa+zc$

and the subspace $W=span(a_1,a_2), a_1=(0,1,0)$ and $a_2=(0,0,1)$

And I'm asked to find the orthogonal projection of $u=(1,0,0)$ in respect of $f$

Here's what I've done so far:

I found an orthonormal basis of $W$ in respect of $f$ using Gram-Schmidt which is

$B=[{\frac{\sqrt2}{2}}(0,1,0),2(0,-\frac{1}{2},1)]$

Now to find the orthogonal projection I tried to calculate:

$pr_wu=pr_{u_1}u+pr_{u_2}u$= $\frac{f[(1,0,0),\frac{\sqrt2}{2}(0,1,0)]}{f[\frac{\sqrt2}{2}(0,1,0),\frac{\sqrt2}{2}(0,1,0)]}\frac{\sqrt2}{2}(0,1,0)+$$\frac{f[(1,0,0),2(0,-\frac{1}{2},1)]}{f[2(0,-\frac{1}{2},1),2(0,-\frac{1}{2},1)]}2(0,-\frac{1}{2},1)$

but $f[(1,0,0),\frac{\sqrt2}{2}(0,1,0)]=0$ and so is $f[(1,0,0),2(0,-\frac{1}{2},1)]$

so in the end I get that $pr_wu=0$

What do I do wrong?

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    $\begingroup$ I think the projection should be zero since under the inner product provided, $u$ is orthogonal to both $a_1$ and $a_2$ $\endgroup$
    – JonHales
    Jun 14 '18 at 19:53
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$$ f[(0,1,0),(0,-1,2)]=-2+2=0 \\ f[(0,1,0),(0,1,0)] = 2 \\ f[(0,0,1),(0,0,1)] = 1 $$ So, you are correct that $\{\frac{1}{\sqrt{2}}(0,1,0),(0,0,1)\}$ is an orthonormal basis of $W$. Therefore, the orthogonal projection of $(1,0,0)$ onto $W$ is $$ \frac{1}{2}f[(1,0,0),(0,1,0)](0,1,0)+f[(1,0,0),(0,0,1)](0,0,1)=(0,0,0). $$ Your answer looks correct to me. This means that $(1,0,0)$ is already orthogonal to $W$. And that can be verified directly, too.

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  • $\begingroup$ Hmm I see thank you very much! $\endgroup$
    – VakiPitsi
    Jun 14 '18 at 20:41

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