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The spectral representation for a self-adjoint operator $T \in L(H)$ for H a Hilbert space is written as: $$ T = \sum_{\lambda \in \sigma(T)} \lambda \pi_{\lambda}, $$ where $\sigma(T)$ is the spectrum of the operator and $\pi_{\lambda}$ is the orthogonal projection on the eigenspace of $\lambda$. This definition is rather vague to me, especially the meaning of $\pi_{\lambda}$ and the way to calculate this in practice. I have been working on an example. $T: l^2 \rightarrow l^2$ defined by: $$ T e_{2k-1} = -\frac{e_{2k}}{\sqrt{k}} $$ $$ T e_{2k} = 2\frac{e_{2k-1}}{\sqrt{k}}. $$ I am supposed to calculate the spectral representation of T*T. I have calculated the adjoint of T to be: $(T^*x)_{2l} = (2/\sqrt{l}) x_{2l-1}$ and $(T^*x)_{2l-1} = (-1/\sqrt{l})x_{2l}$. Now I don't have a clue how to calculate $\pi_{\lambda}$.

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    $\begingroup$ Have you computed $T^*T$? $\endgroup$ – Davide Giraudo Jan 14 '13 at 10:59
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    $\begingroup$ Yes: $(T^*Tx)_{2k} = \frac{4x_{2k}}{k}$ and $(T^*Tx)_{2k-1} = \frac{x_{2k-1}}{k}$ $\endgroup$ – Funzies Jan 14 '13 at 11:06
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    $\begingroup$ So $T^*T$ is a diagonal operator. $\endgroup$ – Davide Giraudo Jan 14 '13 at 11:09
  • $\begingroup$ I'm sorry, but I don't know what that means for $\pi_{\lambda}$. Could you be a bit more specific? $\endgroup$ – Funzies Jan 14 '13 at 11:35
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    $\begingroup$ @Erik you are right $\endgroup$ – Norbert Jan 15 '13 at 9:47

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