Diagonalization and finding orthogonal projection operator

Question

I am struggling with this Linear Algebra course. I attempted these questions but am not confident in my answers. Can someone verify?

Let W = span{1 + x + x$^2$,−x + x$^2$} be a subspace of P2(R) Let us define the inner product <·,·> : V x V → R through = p(−1)q(−1) + p(0)q(0) + p(1)q(1)

Find the orthogonal projection operator [PW]$^a_a$ where α is the standard basis for P2.

Diagonalize [PW]$^a_a$

EDIT: My Incorrect Solution what is wrong with this approach?

W$_1$ = $a_1$ = 1

W$_2$ = $a_2$ - $\frac{(<1+x, 1>)}{(<1, 1>)}$ (1)

W$_2$ = x

W$_3$ = 1 + x + x$^2$ - $\frac{(<1 + x + x^2, x>)}{<x, x>}$(x) - $\frac{(<1 + x + x^2, 1>)}{<1, 1>}$(1)

W$_3$ = -2/3 + x$^2$

I used the Gram-Schmidt method for finding the orthogonal projection operator and came up with the matrix being

$\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\-2/3 & 0 & 1\end{bmatrix}$

For the diagonalization part I found the eigenvalues of this matrix to be 1 with multiplicity of 3. Which would mean the eigenspace would be made up of 3 vectors which I compiled of it being the span{(1, 0, 0), (-1, 1, 0) and (-1, 0, 1)}.

I really think the last part is incorrect but not sure where I made an error in finding [PW]$^a_a$

Thanks!

Answer 1

So $W=\mathrm{span}\{1+x+x^2, -x+x^2\}$. Let me choose a slightly different vectors to represent this space: let $f_1=(x^2-x)/2$ and $f_2=(x^2+3x+2)/(2\sqrt{10})$.

Edit: How did I come up with $f_1, f_2$? The idea is to start with one of vectors, I chose $x^2-x$, and let $f_1=a(x^2-x)$ be such that $\langle f_1, f_1\rangle =1$. Now, for the next part, let $f_2=A(x^2-x)+B(x^2+x+1)$. We have two conditions: $\langle f_1, f_2\rangle = 0$ and $\langle f_2, f_2\rangle = 1$. By doing some algebra, you find $A,B$ and as a result you find $f_2$. This is actually exactly what Gram-Schmidt process does, I just cut-out the middleman.

It is easy to see that $W=\mathrm{span}\{f_1, f_2\}$ but furthermore $\langle f_1, f_1\rangle = \langle f_2, f_2\rangle =1$ and $\langle f_1, f_2\rangle = 0$, i.e. $f_1, f_2$ is an orthonormla basis for $W$.

Now the projection operator $P_W$ sends $v\in P_2(\mathbb{R})$ to $$ \begin{aligned} v\mapsto \langle v, f_1\rangle f_1+\langle v, f_2\rangle f_2 \end{aligned} $$ you can see why it is crucial that 1) both $f_1, f_2$ are actually unit (otherwise $f_1$ is not sent to itself, but a multiple) 2) $f_1, f_2$ have to be orthogonal. In particular $1\mapsto (8+2x+14x^2)/20$ and $x\mapsto (6+19x-7x^2)/20$ and $x^2\mapsto (6-x+13x^2)/20$, making the matrix $$ P_W=\frac{1}{20}\begin{pmatrix} 8 & 6 & 6\\ 2 & 19 & -1\\ 14 & -7 & 13 \end{pmatrix} $$ Once you diagonalize this matrix, it becomes $\mathrm{diag}(1,1,0)$, as it should be. Actually the orthonormal basis $f_1, f_2$ can be extended by $f_3=(-7 x^2-x+6)/(2\sqrt{10})$ making a full orthonomal basis of $P_2(\mathbb{R})$ in which the projection operator is diagonal.

Now, I don't have all of the information about your calculation. But what you found is not the projection matrix. Note that I have essentially used the Gram-Schmidt method in finding my $f_1,f_2$ too. I'm not sure whether you actually checked that your vectors should be unit or not, however.

Answer 2

Your computations for the Gram-Schmidt process were correct, but you applied it to the wrong vectors. To find the projection operator onto a subspace $W$ using G-S, you need to compute an orthogonal basis for that subspace, but you computed one for the entire enclosing space. Even that process as you’ve shown it is a bit puzzling to me. I can’t see where the second vector $1+x$ came from, since it’s not one of the standard basis vectors, nor one of the given spanning vectors of $W$.

What you should’ve done is applied G-S to the set $\{1+x+x^2,x^2-x\}$. It doesn’t matter which of the two vectors you take as the first one, but it looks like the computation might be a little simpler with the second one. So, $w_1=x^2-x$ and $$w_2=(1+x+x^2)-{\langle x^2-x,1+x+x^2\rangle\over\langle x^2-x,x^2-x\rangle}(x^2-x)=\frac12(2+3x+x^2).$$ You’ll also need $\langle w_1,w_1\rangle=4$ and $\langle w_2,w_2\rangle=10$.

From here, you can construct the projection matrix for the operator $\pi_W$ as Hamed describes: $$\pi_W[p]={\langle p,w_1\rangle\over\langle w_1,w_1\rangle}w_1+{\langle p,w_2\rangle\over\langle w_2,w_2\rangle}w_2$$ and the columns of a transformation matrix are the images of the basis vectors, so compute $\pi_W$ for each of the standard basis vectors.

As far as diagonalizing the resulting matrix goes, the eigenvalues of a projection are 0 and 1, and the eigenspace of 1 is the subspace onto which you’re projecting. This means that you’ve already got two of the three eigenvectors that you need, so all you have to do is find an eigenvector of 0, i.e., compute the kernel of the projection matrix.