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Thread: 2 norm of a projector

  1. #1
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    2 norm of a projector

    Let P \in \textbf{C}^{m\times m} be a projector. We prove that \left\| P  \right\| _{2}\geq 1 , with equality if and only if P is an orthogonal projector.

    I suppose we could use the formula \left\| P  \right\| _{2}= max_{\left\| x \right\| _ {2} =1} \left\| Px  \right\| _{2}

    and use the fact that P^{2}=P

    But I am not sure how to proceed.
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  2. #2
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    Quote Originally Posted by math8 View Post
    Let P \in \textbf{C}^{m\times m} be a projector. We prove that \left\| P  \right\| _{2}\geq 1 , with equality if and only if P is an orthogonal projector.

    I suppose we could use the formula \left\| P  \right\| _{2}= max_{\left\| x \right\| _ {2} =1} \left\| Px  \right\| _{2}

    and use the fact that P^{2}=P

    But I am not sure how to proceed.
    Geometrically, there is a good way to see this. Translating it into algebra is not as easy.

    For any vector x\in\mathbb{C}^m, you can write x = Px + (x-Px). This is the unique expression for x as the sum of a vector in the range of P (namely Px) and a vector in the kernel of P ( namely xPx). The projection P is orthogonal if and only if its range and kernel are orthogonal subspaces of \mathbb{C}^m.

    If P is orthogonal then you can apply Pythagoras' theorem to deduce that \|x\|^2 = \|Px\|^2 + \|x-Px\|^2 \geqslant \|Px\|^2. So \|Px\|\leqslant\|x\| (for all x), which tells you that \|P\|=1.

    If P is not orthogonal then you want to find a vector x such that \|Px\|>\|x\|. The way to do that is to take x to be a unit vector in the orthogonal complement of the kernel of P. That's the hard part of the problem, and I don't have time now to explain how the argument goes.
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  3. #3
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    For the first part, can we say that since P is a projector (assuming P \neq 0 ), there is a vector  v \neq 0 such that Pv=v.

    Now take x = \frac{v}{\left\|v \right \|}, so  \left\|Px \right \|_{2} = 1 and using the definition of  \left\|P \right \|_{2} , we get the inequality ?
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