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Let $H$ be a separable Hilbert space and let $T:H\to H$ be a continues linear map such that there exists an orthonormal basis of $H$ that consists of the eigenvectors of $T$. Show that $T$ is normal. That is $T^*T = TT^*$

Any hints would be appriciated!

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up vote 1 down vote accepted

Fix $\{v_j\}$ an orthonormal basis of eigenvectors of $T$. We have $Tv_k=\lambda_kv_k$, where $\lambda_k$ is the corresponding eigenvalue. This gives for each $j$, $$\langle T^*v_j,v_k\rangle=\langle v_j,Tv_k\rangle =\bar \lambda_k\delta_{j,k}.$$ As the sequence $\{v_j\}$ spans a dense subspace, we have that $T^*$ is completely determined. By boundedness and linearity, we just need to show the relationship $$\forall k, TT^*v_k=T^*Tv_k.$$

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$TT^*v_k = T\bar{\lambda_k}v_k = \|\lambda\|^2 v_k = T^*{\lambda_k}v_k =T^*Tv_k $ So then we can use the boundedness and linearity of the composition of maps? – Johan Dec 9 '12 at 16:24
Yes, in other to establish the equality for linear combinations, then for all the vectors. – Davide Giraudo Dec 9 '12 at 16:29
I made the post CW, as the ideas were in @Norbert's answer. – Davide Giraudo Dec 16 '12 at 14:53

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