We consider the Hilbert space with standard inner product, $0<\phi<\frac{1}{2} \pi$. We consider the projection P: $ P \begin{pmatrix} x_1\\ x_2 \end{pmatrix} = \begin{pmatrix} x_1-x_2cot(\phi)\\ 0 \end{pmatrix} $
I need to show that $||P||=\frac{1}{sin(\phi)}$ which I tried by calculating $<Px,Px>$ but I couldn't figure it out. (I got $x_1-x_2cot(\phi)$ again)
But I'm really unsure about my reasoning here. Any help for my problems is very much appreciated :)
$\|P\|=\sup \{|x_1-x_2\cot \phi |\}: x_1^{2}+x_2^{2} \leq 1\}=\sqrt {1+(\cot \phi )^{2}}=\frac 1 {\sin \phi}$. [This is a consequence of Cauchy-Schwarz inequality]. You have already found the matrix of $P^{*}$. You can write $P^{*}(x_1,x_2)=(x_1,-x_1\cot \phi)$.
I have used the following: $\sup \{ax+by: x^{2}+y^{2} \leq 1\} \leq \sqrt {a^{2}+b^{2}}$ by Cauchy Schwarz inequality and the sup is attained when $x=\frac a {\sqrt {a^{2}+b^{2}}}, y=\frac b {\sqrt {a^{2}+b^{2}}}$ Hence, $\sup \{ax+by: x^{2}+y^{2} \leq 1\} = \sqrt {a^{2}+b^{2}}$.