# Finding the orthogonal projection of an inner product space upon a subspace

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.

• 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 $\Bbb R^3$ Aug 28 '19 at 0:51
• 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 :) Aug 28 '19 at 1:12

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 $$U$$.
• 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). Aug 28 '19 at 0:27
• What's wrong is that I got a projection upon $\mathbb R^3$ instead of only $U$ Aug 28 '19 at 0:32
• 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 $U$. Aug 28 '19 at 0:54
• 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. Aug 28 '19 at 1:15
$$(1,2,3)-(1,2,1) = (0,0,2)$$ 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 $$P_U(x,y,z) = \langle (x,y,z),(0,0,1)\rangle (0,0,1) + \frac 15\langle (x,y,z),(1,2,0)\rangle (1,2,0) = \frac 15(x+2y,2x+4y,5z).$$