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Let $H=(\mathbb R^2,(.,.))$ and $M=\{(x,0)|x\in\mathbb R\}, N=\{(x,x\tan(\theta)|x\in\mathbb R)$ with $\theta\in(0,\frac{\pi}{2})$.

Now I would like to find a $T_\theta\in B(H,H)$ with $T^2_\theta=T_\theta, T_\theta(H)=M$ and $Ker(T_\theta)=N$

$B(H,H)$ are the bounded linear operators.

What I am also intersted in is how to calculate $||T_\theta||$ then, but I have no idea finding the correct $T_\theta$

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Try projections – Hui Yu Nov 28 '12 at 2:42
Could you explain it a little bit more? – Voyage Nov 28 '12 at 7:05
What is the projection onto M along N? – Hui Yu Nov 28 '12 at 8:39
It should be calculated somehow by using the kernel and the image. My problem is to write it down explicitly. Let $P$ the projection onto M along N and $h\in H$ => $T_\theta h=Ph$ But what is $T_\theta$ then? – Voyage Nov 28 '12 at 13:27
$P_M(h)=\sum_{i=1}^{k}\frac{<h,m_i>}{<m_i,m_i>}m_i$ where m is an orthogonalbasis of M ? – Voyage Nov 28 '12 at 13:35
up vote 0 down vote accepted

I write down briefly the ideas. $T^2=T$ contains a lot of information, but what is most useful here is that

If $x\in\operatorname{Range}(T)$ then $Tx=x$.

So $T$ fixes everything in the range, and in this case \begin{equation} Te_1=e_1, \end{equation}so the first column of $T$ is $[1;0]$.

I guess then you can pretty much solve $T$ by a system of linear equations.

Or, by noting that everything in $M$ is an eigenvector of $T$ with eigenvalue $1$ (because they are fixed by $T$), and everything in $N$ is an eigenvector of $T$ with eigenvalue $0$ (they are annihilated by $T$).

I guess in $\mathbb{R}^2$ knowing two eigenvalues and corresponding eigenvectors gives everything you need to know about $T$.

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Thank you, is it possible that the second column of T is $[-1/(tan(\phi));0]$ – Voyage Nov 28 '12 at 14:57
I guess not :). It should be a linear relation between $e_1$ and $e_2$. – Hui Yu Nov 28 '12 at 15:06
Well, but T={{1,c},{0,d}} and $[x;x\tan(\phi)]$ is an eigenvector of T with eigenvalue 0 => $T*[x;x\tan(\phi)]$=0 ?? How does the system of linear equation you first mentioned looks like? – Voyage Nov 28 '12 at 15:26

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