I'm studying for a qualifying exam in algebra and I'm slightly stuck on the following problem:
Let $\textsf V$ be an $n$-dimensional vector space over $\mathbb{R}$ with $n\geq1$ endowed with an inner product $\langle \cdot , \cdot \rangle$, and let $\textsf{T}: \textsf{V}\to \textsf{V}$ be a linear transformation.
Prove that the following statements are equivalent:
$(\textrm{i})$ For all $v\in \textsf V$, $\langle \textsf{T}(v),\textsf{T}(v)\rangle=\langle v,v \rangle$.
$(\textrm{ii})$ For all $u,v\in \textsf V$, $\langle \textsf{T}(u),\textsf{T}(v)\rangle=\langle u,v\rangle$.
$(\textrm{iii})$ There is some nonzero $w\in \textsf V$ such that $\langle \textsf{T}(w),\textsf{T}(w)\rangle=\langle w,w\rangle$, and for $u,v\in \textsf V$ we have $\langle \textsf{T}(u),\textsf{T}(v)\rangle=0$ if and only if $\langle u,v\rangle=0$.
I'm pretty sure this is a simple problem, but I'm just not seeing it. Clearly $(\textrm{ii})$ implies $(\textrm{i})$ and $(\textrm{iii})$ trivially, but I don't see any other clear implications.
Maybe I need to use the fact that this is a real vector space to exploit conjugate symmetry? This almost has the flavor of some self-adjoint property, but it seems simpler than that.
i) implies ii) follows from the identity $\langle x , y \rangle =\frac {\langle (x+y) , (x+y) \rangle -\langle (x-y) , (x-y) \rangle} 2$. ii) implies iii) is obvious
For iii) implies i) take an orthonormal basis $(u_i)$. Then, for $i \neq j$ we have $\langle u_i , u_j \rangle =0$ and iii) implies $\langle T(u_i) ,T( u_j) \rangle =0$. Next observe that $\langle (u_i-u_j) , (u_i+u_j) \rangle =0$ which gives $\langle T(u_i-u_j) , T(u_i+u_j) \rangle =0$ . Simplify this to conclude that $\langle Tu_i , Tu_i \rangle =\langle Tu_j , Tu_j \rangle$. Hence $\langle Tu_i , Tu_i \rangle$ is a number $a$ independent of $i$. Now use the fact that any vector $u$ can be expanded as $\sum c_iu_i$ to prove i). I leave the details of this last part to you since it is quite routine.