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2-norm of a projector is greater than 1

  1. May 6, 2010 #1
    Let [tex]P \in \textbf{C}^{m\times m}[/tex] be a projector. We prove that [tex]\left\| P \right\| _{2}\geq 1[/tex] , with equality if and only if P is an orthogonal projector.

    I suppose we could use the formula [tex]\left\| P \right\| _{2}= max_{\left\| x \right\| _ {2} =1} \left\| Px \right\| _{2}[/tex] and use the fact that [tex]P^{2}=P[/tex] and [tex]P=P^{*}[/tex] (P* is the transpose conjugate of P).

    But I am not sure how to use these.
  2. jcsd
  3. May 6, 2010 #2


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    Technically this isn't true since [itex]P = 0[/itex] is a projection matrix with [itex]||P||_2 = 0[/itex]. But assuming [itex]P \neq 0[/itex], there exists some [itex]v \neq 0[/itex] such that

    [tex]Pv = v[/tex]

    If you use this fact along with the definition of the 2-norm of a matrix, your inequality follows pretty much immediately.
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