Orthogonal projection and bases

Question

Let $P: \mathbb R^3\to \mathbb R^3$ is orthogonal projection on $x\mathbb Oy$ plane:

a)Prove that is linear transformation

b)Find a matrix of $P$ using standard base.

c)Find matrix of that linear transformation using base $B=\{(1,0,1,),(1,1,0),(1,1,1)\}$

For a) I know, but for b) that is $P(e_1)=e_1$, $P(e_2)=e_2$, $P(e_3)=0$, so matrix P is

$\begin{bmatrix} 1 &0 &0\\ 0& 1& 0\\ 0& 0& 0 \end{bmatrix}$

and for c)

If I use formula $P=S^{-1}P´S,$

where

$S^{-1}=\begin{bmatrix} 1& 1& 1\\ 0& 1& 1\\ 1& 0& 1 \end{bmatrix}$

$S=\begin{bmatrix} 1& -1& 0\\ 1& 0& -1\\ -1& 1& 1 \end{bmatrix}$

so $SPS^{-1}=P´$ ¨ then $P´=\begin{bmatrix} 1& 0& 0\\ 1& 1& 1\\ -1& 0& 0 \end{bmatrix}$

is this ok?

Answer 1

It seems all correct, indeed also for point "c" we have that

$$S^{-1}=\begin{bmatrix} 1& 1& 1\\ 0& 1& 1\\ 1& 0& 1 \end{bmatrix}$$

represents the change of basis from the new basis to the standard one therefore we have

$$P'=SPS^{-1}$$

Answer 2

Almost OK, but it's rather with $S$ and $S^{-1}$ exchanged.
If $B$ denotes the matrix with the new basis as column vectors, $PB$ will produce their images under $P$, coordinated in the standard basis, and $B^{-1}$ of this will turn them to coordinates in the new basis.

Answer 3

Well, I haven't seen the notation $x\mathbb Oy$ before, but I guess I can figure it out from context.

For part (c), you can you change of basis, or you can just work it out from its action. Let $B=\{b_1,b_2,b_3\}$. Then $P(b_1)= (1,0,0)= b_1+b_2-b_3$. So $P(b_1)$, written in terms of the basis $B$, is $(1,1,-1)$. $P(b_2)=0b_1+b_2+0b_3 \rightarrow (0,1,0)$. $P(b_3)$ is also just $b_2$, or $(0,1,0)$. So to create the matrix representation of $P$, we just take those as columns vectors:

$P=\begin{bmatrix} 1& 0& 0\\ 1& 1& 1\\ -1& 0& 0 \end{bmatrix}$

Note that there's a difference between "the matrix of P" and "the matrix P". $P$ is not a matrix, it's a linear operator. In the standard basis, it's represented by the matrix

$\begin{bmatrix} 1 &0 &0\\ 0& 1& 0\\ 0& 0& 0 \end{bmatrix}$

but as this question shows, $P$ is represented by a different matrix in a different basis, so we need to distinguish between the two.