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Let $V$ be a complex inner product space with $\dim(V)<\infty$. Let $T$ be a normal operator ($TT^*=T^*T$). Show that $V$ is the direct sum of $\text{null}(T)$ and $\text{range}(T)$. Show that for any $S$, if $ST=TS$, then $ST^{*}=T^{*}S$.

For the first one, we know that the rank-nullity theorem

$$\dim(V) = \dim( \text{null}(T) ) + \dim( \text{range}(T) )$$

Is it enough to say that "$V$ is the direct sum of $\text{null}(T)$ and $\text{range}(T)$"?

Also, for the second one. I am not how to the first result to prove that. I found this is not a trivial result in functional analysis (called Fuglede's theorem.)

Arctic Char
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Hermi
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  • The statement "$V$ is the direct sum of $\text{null}(T)$ and $\text{range}(T)$" means $V = \text{null}(T) \oplus \text{range}(T)$. And the proof of $ST=TS$ implies $ST^=T^S$ is in your link, isn’t? Write $T=\sum_i\lambda_iP_i$ and note that $ST=TS$ iff $SP_i=P_iS$ for each $i$. Hence, $S$ commutes with $T^*=\sum_i\overline{\lambda_i}P_i$. – azif00 Mar 12 '23 at 05:47
  • This has been asked on this site but I cannot locate the duplicate. There is no need to mess with the null space and image of $T$. Briefly speaking, since $\lambda\mapsto\overline{\lambda}$ is injective, $T^\ast$ can be written as a polynomial in $T$ (using e.g. Lagrange interpolation). Now the result is trivial. – user1551 Mar 12 '23 at 08:35
  • Does this answer your question? Fuglede's theorem in finite-dimensional vector space – user8675309 Mar 12 '23 at 16:59
  • I marked this as a duplicate of prior Fuglede's Theorem question(s). The first part immediately follows when you show $\text{rank}\big(T^2\big) =\text{rank}\big(T\big)$. And if you know basics about Hermitian (finite dim) Operators then you can write $\text{rank}\big((T^T)^2\big) =\text{rank}\big((T^)^2T^2\big)\leq \text{rank}\big(T^2\big)\leq\text{rank}\big(T\big)=\text{rank}\big((T^T)\big)=\text{rank}\big((T^T)^2\big)$ i.e. these inequalities are met with equality so part 1 is done. – user8675309 Mar 12 '23 at 17:04
  • @user8675309 why the first part follows from $rank(T^2)=rank(T)$? – Hermi Mar 12 '23 at 19:48
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    You always have $\text{rank}(T^2)\leq \text{rank}(T)$. This is met with equality iff $T$ is injective on the vector space $\text{im }T= (T\cdot V)$. Prove this to yourself by supposing $Tw = \mathbf 0$ for some nonzero $w\in T\cdot V$ and extending to a basis for $T\cdot V\implies \text{rank}(T^2)\lt \text{rank}(T)$. Now $\dim\big(\text{im }T+\ker T\big)=\dim\big(\text{im }T\big)+\dim\big(\ker T\big)+\dim\big(\ker T\cap\text{im }T \big)$ $=\dim\big(\text{im }T\big)+\dim\big(\ker T\big)+0 = \dim V\implies V=\text{im }T+\ker T$. Your post needed to deal with the intersection. – user8675309 Mar 13 '23 at 04:52

1 Answers1

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Fix the $T$, the result is certainly true for diagonalisable matrices. now the set of $V(ST-TS) = \{S\in \mathbf M_n(\mathbb C):ST-TS=0\}$ and $V(ST^*-T^*S) = \{S\in \mathbf M_n(\mathbb C):ST^*-T^*S=0\}$ are two closed complex analytic variety. From Schur's theorem, it is clear that the set of diagonalisable matrices is able matrices are dense in $\mathbf M_n(\mathbb C)$. Let $i$ and $i'$ be their closed topological immersion into $\mathbf M_n(\mathbb C)$, which agree on the intersection with $N$, which is still dense the two subspace respectively. Since immersion is continuous and the space is Hausdorff, the two immersions agree completely.

bocchi_the_form
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    "From Schur's theorem, it is clear that the set of normal matrices are dense in $M_n(\mathbb C)$" can't be true. You may want to read https://math.stackexchange.com/questions/4646701/density-of-normal-matrices-in-m-n-mathbbc – user8675309 Mar 13 '23 at 04:58