Question: q(x1..............  Edit

Answer: Suppose ⟨Tv,v⟩>0for every v∈V with v≠0. If T is not invertible, there must exist a nonzero u∈V such that Tu=0, hence ⟨Tu,u⟩=0 for u≠0. Therefore we get a contradiction, which implies that Tis invertible.

Conversely, suppose Tis invertible. Since T is positive, it follows from 7.35 (a) ⟺ (d) that there exists a self-adjoint operator S such thatthat S^2=T. Because T is injective, so is S. Hence for every v∈V with v≠0, we have Sv≠0. Moreover, since S is self-adjoint (and Sv≠0), we have
⟨Tv,v⟩=⟨S^2v,v⟩=⟨Sv,S*v⟩= ⟨Sv,Sv⟩>0.

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