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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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    $P_M(h)=\sum_{i=1}^{k}\frac{}{}m_i$ where m is an orthogonalbasis of M ?2012-11-28

1 Answers 1

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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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    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?2012-11-28