Let be $U=${$(x,y,z) \in \mathbb{R}^3 | x-2y+3z=0$} Find $P_u(1,1,1)$
This is what i did.
If $(x,y,z) \in U$ then $x=2y-3z$. That can be expressed by $(x,y,z)=(2y-3z,y,z)=(2,1,0)y+(-3,0,1)z$
So $U=<(2,1,0),(-3,0,1)>$
Using Gram-Schmidt and the orthonormal form.
$w_1=\frac{1}{\sqrt{5}}(2,1,0)$
$w_2=\frac{5}{\sqrt{70}}(\frac{-3}{5}, \frac{6}{5},1)$
So $A=\begin{bmatrix}
\frac{2}{\sqrt{5}} & -\frac{3}{\sqrt{70}}\\
\frac{1}{\sqrt{5}} & \frac{6}{\sqrt{70}}\\
0 & \frac{5}{\sqrt{70}}\\
\end{bmatrix}
$
And using this: $P=A(A^TA)^-A^T$
Then $P=\begin{bmatrix}
\frac{13}{14} & \frac{1}{7} & -\frac{3}{14}\\
\frac{1}{7} & \frac{5}{7} & \frac{3}{7}\\
-\frac{3}{14} & \frac{3}{7} & \frac{5}{14}\\
\end{bmatrix}
$
So $P_u=(\frac{6}{7}, \frac{9}{7},\frac{4}{7})$
Am i doing this right? I'm currently learning this in class.
Best Answer
The answer is correct and the method you used is sound, although I’d say that it’s overkill for this particular problem. If you were going to project a lot of vectors, then finding the projection matrix would be worthwhile, but you can compute the projection of a single vector with a lot less work.
First, if you’re going to compute the projection matrix via the formula $P=A(A^TA)^{-1}A^T$, then there’s no reason to go through the Gram-Schmidt process to compute an orthonormal basis for $U$. This formula works for any basis. So, once you had your initial basis for $U$, you could’ve simply applied the formula to $A=\begin{bmatrix}2&1&0\\-3&0&1\end{bmatrix}^T$ instead.
On the other hand, once you’ve gotten an orthonormal basis for $U$, you can take advantage of the fact that the projection onto $U$ is the sum of the individual projections onto those basis vectors. So, setting $v=(1,1,1)^T$, compute ${w_1^Tv\over w_1^Tw_1}w_1+{w_2^Tv\over w_2^Tw_2}w_2$. If you need the matrix $P$ for some reason, then it is ${w_1w_1^T\over w_1^Tw_1}+{w_2w_2^T\over w_2^Tw_2}$, which is a lot easier to calculate than the expression in the previous paragraph. (Note the order of the multiplicands in the numerators: those aren’t inner products. They’re the outer products of a vector with itself, which is a $3\times3$ matrix.).
Since $U$ is two-dimensional, the easiest way by far to compute a projection onto $U$ is to compute the orthogonal rejection, which is the projection onto a vector orthogonal to $U$, and then subtract that from the original vector. In this case, we read from the defining equation of $U$ that $w=(1,-2,3)^T$ is orthogonal to $U$, so the projection of $v=(1,1,1)^T$ onto $U$ is $$v-{w^Tv\over w^Tw}w=(1,1,1)^T-\frac2{14}(1,-2,3)^T=\left(\frac67,\frac97,\frac47\right)^T,$$ just as you got.