Linear bounded operator T on $l^2$ is given by : $T(x_1,x_2,x_3,..) := (x_1,x_1,x_1,x_2,x_2,x_2,x_3,x_3,x_3,..)$.
Prove that $\frac{T^*T}{3}$ and $\frac{TT^*}{3}$ are orthogonal projections.
Thanks in advance for the help
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You have $$ \langle T^*Tx,y\rangle=\langle Tx,Ty\rangle= 3\sum_nx_n\overline{y_n}=\langle 3x,y\rangle. $$ So $T^*T=3I$. Since $\ker TT^*=\ker T^*$, we have $$ \overline{\text{ran}\,TT^*}=\ker(TT^*)^\perp=\ker (T^*)^\perp=\overline{\text{ran}\,T}. $$ So the range of $TT^*/3$ equals the range of $T$. And it is a projection, since $$ \left(\frac{TT^*}3\right)^2=\frac{TT^*TT^*}9=\frac{3TT^*}9=\frac{TT^*}3. $$ So $TT^*/3$ is the orthogonal projection onto the range of $T$.
Martin Argerami
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Just write the inner product. – Martin Argerami Feb 12 '18 at 23:17
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And for $TT^$ I should calculate $<T^x,T^*y>$ ? – David Feb 13 '18 at 02:07
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Can you please help with $TT^*$ ? Thanks in advance – David Feb 13 '18 at 02:16
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What for? You wanted to show that $TT^*/3$ is an orthogonal projection, and I gave you a proof of that. – Martin Argerami Feb 13 '18 at 02:54
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And for showing that $T^T$ is orthogonal projection is enough to say that $T^T$ = $3I$? – David Feb 13 '18 at 03:03
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$T^T$ is not an orthogonal projection. It's $T^T/3$, that is equal to the identity. The identity is an orthogonal projection. – Martin Argerami Feb 13 '18 at 03:14
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And because range of range $\frac{TT^*}{3}$ = range of T and it is a projection we know that it is orthogonal projection? – David Feb 13 '18 at 16:37
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No, it's because it is selfadjoint. – Martin Argerami Feb 13 '18 at 18:17
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Directly from the matrices (the transpose conjugate of that of $T$) you get that $$T^*(x_1,x_2,x_3,...)=(x_1+x_2+x_3,x_4+x_5+x_6,...)$$
Therefore $$\frac{1}{3}T*T(x_1,x_2,x_3,...)=(x_1,x_2,x_3,...)$$
and
$$\frac{1}{3}TT*(x_1,x_2,x_3,...)=(\frac{x_1+x_2+x_3}{3},\frac{x_1+x_2+x_3}{3},\frac{x_1+x_2+x_3}{3}, \frac{x_4+x_5+x_6}{3},\frac{x_4+x_5+x_6}{3},\frac{x_4+x_5+x_6}{3},...)$$
which projects over the space generated by $(1,1,1,0,...), (0,0,0,1,1,1,0,...),...$