Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

up vote 5 down vote accepted

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

share|improve this answer
    
Oh it was this easy! thanks! –  Johan Dec 13 '12 at 12:44
add comment

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 $$ A^* A = \begin{bmatrix} 0 & 0 \\ 0 & 2 \end{bmatrix}$$

share|improve this answer
    
thanks good example!!! –  Johan Dec 13 '12 at 12:43
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.