So I'm confused about projections. Following from my question here
What is the difference between this question
$1)$ "Find the standard matrix of the linear operator $T:R^2\rightarrow R^2$ given by the orthogonal projection onto the vector $(1,-2)$."
vs
$2)$ "Let $T:R^3\rightarrow R^3$ be the orthogonal projection onto the place $3x+4z=0.$ Find the standard matrix $[T]$, of $T$.
In the first question, I would go:
$T(e_1)=\frac{(1,0)\cdot(1,-2)}{\|(1,-2)\|}(1,-2)$
$=(\frac{4}{5},\frac{2}{5})$
and
$T(e_2)=\frac{(0,1)\cdot(1,-2)}{\|(1,-2)\|}(1,-2)$
$=(\frac{-2}{5},\frac{4}{5})$
So $[T]$=$\left(
\begin{array}{cc}
\frac{1}{5} & \frac{-2}{5}\\
\frac{-2}{5} & \frac{4}{5} \\
\end{array}
\right)
$
However for the 2nd question, it's similar but:
The normal of this plane is $<3,0,4>$. So:
$T(e_1)=(1,0,0) – \frac{(1,0,0)\cdot(3,0,4)}{\|(3,0,4)\|}(3,0,4)$
$=(\frac{16}{25},0,\frac{-12}{25})$
Similarly, $T(e_2)=(0,1,0)$ and $T(e_3)=(\frac{-12}{25},0,\frac{9}{25})$
Thus,
$[T]$=$\left(
\begin{array}{ccc}
\frac{16}{25} & 0 & \frac{-12}{25}\\
0 & 1 & 0\\
\frac{-12}{25} & 0 & \frac{9}{25}\\
\end{array}
\right)
$
My question is, in the first question, why did I not have to subtract the projection from the standard basis vectors? What's the difference? Is there a reason I have to do subtraction in the 2nd question?
Best Answer
Take a closer look at what your calculations for the second problem represent. You’re computing the orthogonal projection onto the normal to the plane—pretty much the same thing that you did for the first problem—but then subtracting it from the original vector. That is, in order to compute the orthogonal projection onto a subspace of $\mathbb R^3$, you’re effectively decomposing the ambient space into the direct sum of this subspace (the plane) and its orthogonal complement (the line defined by the plane’s normal). The projection onto the latter is somewhat simpler to compute directly. You could’ve instead found an orthonormal basis for the plane and worked directly with orthogonal projections onto those vectors (or even computed the projection matrix from any basis of the plane), but that way involves a lot more work.