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Show that there are linear operators T on the Hilbert space H what are not orthogonal projections, but their spectrum consists of the eigenvalues $\{0,1\}.$

I can not come up with an counterexample, but I suppose it must be infinite dimension? Hilbert spaces has orthogonal bases and I did an exersice showing that for projections $T'(T-I) = 0$ So obviously $T \neq TT'$, is zero some kind of accumulation point?

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up vote 5 down vote accepted

$$\begin{pmatrix}0 & 0 & 0 \\ 0 & 1 & 57 \\ 0 & 0 & 1\end{pmatrix}$$

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Oh it was this easy! thanks! – Johan Dec 13 '12 at 12:44

To complement Nate Eldredge's example, it might be more interesting to consider operators that are projections, but not orthogonal projections. We can rephrase this in terms of the SVD: An orthogonal projection is an operator $T$ such that $T^2 = T$, and such that the singular values of $T$ are each either $0$ or $1$.

A non-orthogonal projection, also called an oblique projection, is one for which $T^2 = T$, but the singular values of $T$ may not belong to $\{0,1\}$. Consider this example: $$ A = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}$$ We can check that $A^2 = A$, so this is a projection, but it is not an orthogonal projection, since $$ A^* A = \begin{bmatrix} 0 & 0 \\ 0 & 2 \end{bmatrix}$$

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thanks good example!!! – Johan Dec 13 '12 at 12:43

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