Question

Suppose T ∈ L(V ) is normal. Prove that if range T = range T*

Answer 1

T is normal implies that $||Tv||=||T^*v||$ for all v. Thus, if $v \epsilon \space null \space \space T$ then $||Tv||=0$ implies that$||T^*v||=0$ , thus, if $v \epsilon \space null \space \space T^*$. As $(T^∗)^∗ = T$ , this means that $v \epsilon \space null \space \space T$ iff $v \epsilon \space null \space \space T^*$. So the kernels of T and T^∗ are equal.

$Null \space \space T^∗ = (range \space \space T )^{⊥}$ and $Null \space \space T = (range \space \space T^*)^{⊥}$ . As null T = null T^∗, this implies that

$(R ange\space \space T)^{⊥}= (range \space \space T^*)^{⊥}$

If U is a subspace of V, then $(U)^⊥$ = U. Taking the orthogonal complement of both sides of the above equation gives us range T = range T^∗.