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I would like to check if my understanding of projection maps is correct.

I have been given the following subset of $\mathbb{R}^3$:

$$A=\left\{\begin{pmatrix} x \\ y \\ -x+2y \end{pmatrix} \middle| x,y,z\in\mathbb{R}\right\}$$

A basis for this subset is $\mathscr{B}=\left\{ \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} \right\}$, and to extend this basis to one for the vector space $\mathbb{R^3}$ we simply add to the basis the vector: $$\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$$

To obtain $\mathscr{C} = \left\{ \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right\}$, a basis for $\mathbb{R}^3$.

We can call $B = Span\left\{\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\right\}$, and then we can say $\mathbb{R}^3=A\bigoplus B$.

What I want to know is if I am correct in interpreting the definition of projection map. Let $P:\mathbb{R}\to\mathbb{R}$ be the projection map onto A. The question asks me to calculate $P(e_1)$, $P(e_2)$ and $P(e_3)$ then write down the matrix of $P$ with respect to the standard basis of $\mathbb{R}^3$. Without explicitly giving my answer (I want to check my method, not my answers), this is my method:

  1. Write each vector $e_1$, $e_2$ and $e_3$ as a linear combination of the vectors in $\mathscr{C}$, so, for example, $e_1 = \alpha\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}+\beta\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} \gamma\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.

  2. For the projection map onto $A$ we take only the first two terms as the first two terms are in the basis $\mathscr{B}$. So, for the combination in step 1, $P(e_1)=\begin{pmatrix} \alpha \\ \beta \\ \gamma \end{pmatrix} = \alpha e_1+\beta e_2+\gamma e_3$

  3. To form the matrix P we write down the columns of the matrix the coefficients describe in the last step, so we get: $P=\begin{pmatrix} \alpha & . & . \\ \beta & . & . \\ \gamma & . & . \end{pmatrix}$, and fill in the missing columns as we did for the first column above.

Am I correct in my method? If I have any of this wrong, please guide me in the right direction.

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Basically, how to say, you got the most part of it right. But,

  1. you "illustrated" a case by $1,2,-3$ numbers. But.. as I guess, letters in algebra were just presented because of these kind of "illustrations". So, better use letters for unknown numbers. (In the second time, your $-3$ turned into $+3$.) Anyway, those numers are $1,0,-1$.
  2. It is written that $P$ is "the" projection map to $A$. But, there can be more projections to any nontrivial subspace. What you calculated, is also a projection, and the method is good. But probably the orthogonal projection was meant. So, find a third basis element which is orthogonal to $A$, and then you're done. (But, if the exercise was first to show an arbitrary complement subspace $B$, then this $P$ was meant by "the projection").
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Thank you, I fixed your first point. In class we've not defined orthogonal projections, but the question does ask for "the" projection - what are the implications of this? – Jon Oct 24 '12 at 20:09

(This discussion applies to finite dimensional spaces. Recall that to define a linear operator, it is sufficient to define its behaviour on a basis.)

Your approach is correct. However, there is some ambiguity in defining $P$, you have defined one projection, but there are others.

Suppose you have a projection $P$ onto a subspace $A$ with basis $a_1,.., a_k$. Then $P$ is uniquely defined on $A$ (since $Pa_i = a_i$). Suppose $b_{k+1},.., b_n$ together with $a_1,.., a_k$ form a basis for the whole space. Then $P$ can be arbitrarily defined on $b_i$ as long as $P b_i \in A$. So, the projection is not unique.

Back to the problem on hand:

Let $v_1,v_2,v_3$ be the vectors you have in $\mathscr{C}$ which form a basis for $\mathbb{R}^3$ (and $v_1, v_2$ form a basis for $A$). Then you must have $P v_1 = v_1$, $P v_2 = v_2$, but the only requirement for $P v_3$ is that it lie in $A$. So $P v_3 = \alpha_1 v_1 + \alpha_2 v_2$, where $\alpha_i$ are arbitrary (but fixed, of course).

Let $V = \begin{bmatrix} v_1&v_2&v_3 \end{bmatrix}$. Then we have $PV = \begin{bmatrix} v_1&v_2& \alpha_1 v_1 + \alpha_2 v_2\end{bmatrix} = W$.

You were asked to compute $P e_i$, which is tantamount to computing $P = W V^{-1}$. Noting that $W = \begin{bmatrix} v_1 & v_2 & 0 \end{bmatrix} + \alpha_1 \begin{bmatrix} 0 & 0 & v_1 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 0 & v_2 \end{bmatrix}$, we see that P can be expressed as

$$ P = \begin{bmatrix} v_1 & v_2 & 0 \end{bmatrix}V^{-1} + \alpha_1 \begin{bmatrix} 0 & 0 & v_1 \end{bmatrix} V^{-1} + \alpha_2 \begin{bmatrix} 0 & 0 & v_2 \end{bmatrix} V^{-1}$$

Grinding through through the calculation gives: $$P = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 2 & 0 \end{bmatrix} + \alpha_1 \begin{bmatrix} 1 & -2 & 1 \\ 0 & 0 & 0 \\ -1 & 2 & -1 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 0 & 0 \\ 1 & -2 & 1 \\ 2 & -4 & 2 \end{bmatrix}$$

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