Answer to Question #153615 in Linear Algebra for Sourav Mondal

Question

 Let (V, <, >) be an inner product space and

let T belongs to A(V). Prove that the following 

conditions are equivalent. 

(i) T*T =I

(ii) <Tx, Ty> = <x,y> for all x, yEV 

(iii) ||Tx||=||x||for all xEV. 

Answer 1

$(iii)\rightarrow (ii)$

Suppose that $||T(x)|| = ||x||$ for all $x ∈ V$.

Now for any $x, y ∈ V :\;\; <x,y>=\frac{1}{4}\sum_{k=1}^{4}i^k||x+i^ky||^2$

if $\mathbb{F}=\mathbb{C},$

And if $\mathbb{F}=\R$ ,

we have similar result: $<x,y>=\frac{1}{4}\sum_{k=1}^{4}(||x+y||^2-||x-y||^2)$

Using the above relation for the complex version, we have

$<T(x),T(y)>\\=\frac{1}{4}\sum_{k=1}^{4}i^k||T(x)+i^kT(y)||^2\\ =\frac{1}{4}\sum_{k=1}^{4}i^k||T(x+i^ky)||^2\\=\frac{1}{4}\sum_{k=1}^{4}i^k||x+i^ky||^2\\ =<x,y>$

$(ii)\rightarrow (iii)$

Suppose that $<T(x),T(y)>=<x,y>\forall x,y\in V$.

Substituting $y = x,$ we have the desired result. 

$(ii)\rightarrow (i)$

Suppose $T$ is unitary.

Then $T$ has an inverse $T^{-1}$.

So, for $x, y ∈ V$ ,

$<T(x), y> = <T(x), TT^ {−1} (y)> = <x, T^{−1} (y)>$ .

So, $T^*$ exists and $T^*=T^{-1}$

$\Rightarrow TT^*=I$

$(i)\rightarrow (ii)$

Assume the adjoint $T^*$ exists

and $TT^*=T^*T=I$

Now for $x, y ∈ V$, we have $<T(x), T(y)) = <x, T^ ∗T(y)> = <x, y>$

Therefore from above discussion, we can conclude that,

$(iii)\iff(ii)\;\; and \;\;(i)\iff (ii)$

So that implies $(i)\iff(ii)\iff(iii)$

$\therefore$ $\pmb{(i),(ii),(iii)}$ are equivalent.