0
$\begingroup$

Let $V=\mathbb R^3$ be an inner product space with the standard inner product (that means $\langle(x_1,y_1,z_1),(x_2,y_2,z_2)\rangle=x_1y_1+x_2y_2+x_3y_3$ ).
$U=span\{(1,2,3),(1,2,1)\}\subseteq V$

a) Find an orthonormal basis for $U$
b) Find an explicit formula for the orthogonal projection $P_U: V\rightarrow U$

So I found an orthonormal basis for $V$, denote that basis $B=(b_1,b_2)=\bigg(\frac{1}{\sqrt6}(1,2,1),\frac{1}{\sqrt{30}}(-1,-2,5)\bigg)$, and I'm totally sure it's an orthonormal basis, so no problem with that. On the other hand, what I'm having trouble with is finding the explicit formula. What I did was to use the following formula for the orthogonal projection:

Let $W\subseteq V$ be a vector subspace, where $V$ is an inner product vector space of finite dimension, and let $S=\{s_1,\dots,s_k\}$ be an orthonormal basis for $W$, then the orthogonal projection upon $W$ is given by: $P_W(u)=\sum_{i=1}^k\langle u,s_i\rangle\cdot s_i$

This is the way I used it:
$P_U(x,y,z)=\langle(x,y,z),b_1\rangle\cdot b_1+\langle(x,y,z),b_2\rangle\cdot b_2=\dots=\frac{1}{5}(x+2y,2x+4y,5z)$

The problem is that I got a projection upon $\mathbb R^3$, instead of only upon $U$ (I took vectors$\in\mathbb R^3\backslash U$ to find that out), and I dont get what is wrong with what I did.

Any help would be appreciated.

Edit: I found this similar post: orthogonal projection formula question , but since we didn't learned cross products I can't use the solution given there.

$\endgroup$
2
  • 1
    $\begingroup$ when you apply $P_U$ to any element of $\Bbb R^3$, the result has a second component twice the first, which is characteristic of $U$, not all of $\Bbb R^3$ $\endgroup$ Aug 28 '19 at 0:51
  • $\begingroup$ it appears that I did a miscalculation concluding that $P_U$ maps some vectors to $\mathbb R^3$. In that case, should I delete the post? (I'm a new user, and find the rules be sometimes hard to understand) anyway, thanks for your affirmation, I can see I was getting the right answer after all :) $\endgroup$
    – Bernard
    Aug 28 '19 at 1:12
1
$\begingroup$

The range of $P_U$ is two dimensional. How did you conclude that it is whole of $\mathbb R^{3}$? I did not check all your calculations but your method is correct and it is likely that you have obtained the projection correctly.

Edit: I have checked your calculations and everything seems perfect. Your $P_U$ is the projection with range $U$.

$\endgroup$
5
  • $\begingroup$ I didn't conclude that that it's all of $\mathbb R^3$. I got something wrong, but can't put a finger on what is my mistake. I found this similar post: math.stackexchange.com/questions/1809883/… , but I can't use cross products (we didn't leraned it). $\endgroup$
    – Bernard
    Aug 28 '19 at 0:27
  • $\begingroup$ it is difficult to help you unless you spell out exactly what is wrong according to you with the answer you obtained. $\endgroup$ Aug 28 '19 at 0:30
  • $\begingroup$ What's wrong is that I got a projection upon $\mathbb R^3$ instead of only $U$ $\endgroup$
    – Bernard
    Aug 28 '19 at 0:32
  • $\begingroup$ If $f: A \to B$ is a map, and $C \subset B$, and $f(A) \subset C$, then there's a very natural map, $g: A \to C : a \mapsto g(a) = f(a)$ from $A$ to this subset of $B$. You're in that situation: the image of the map you defined lies entirely within $U$, hence you can redefine your map as going from $\Bbb R^3$ to $U$. $\endgroup$ Aug 28 '19 at 0:54
  • $\begingroup$ it appears that I did a miscalculation (took a lin. dependent vector in $U$ when I thought it was from $\mathbb R^3\backslash U$), so I was wrong about $P_u$ mapping to $\mathbb R^3$. Thanks for the answer. $\endgroup$
    – Bernard
    Aug 28 '19 at 1:15
1
$\begingroup$

$(1,2,3)-(1,2,1) = (0,0,2)$ and $(1,2,3) - 3(1,2,1) = (-2,-4,0)$. Hence, $$ U = \operatorname{span}\{(0,0,1),(1,2,0)\}. $$ These vectors are already nicely orthogonal and an orthonormal basis is $\{(0,0,1),\tfrac{1}{\sqrt 5}(1,2,0)\}$. The orthogonal projection is thus given by $$ P_U(x,y,z) = \langle (x,y,z),(0,0,1)\rangle (0,0,1) + \frac 15\langle (x,y,z),(1,2,0)\rangle (1,2,0) = \frac 15(x+2y,2x+4y,5z). $$

$\endgroup$

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .