Let $V$ be a linear vector space with inner product, if $T$ is a operator on $V$, how can i prove that $Dim(Im (T))=1$ iff exists $u, w \neq 0$ in $V$, such that $T(v)=<v,u>w$ for all $v \in V$.
my try:
if $Dim(Im (T))=1$ , then there is $w$ that generate the image. Then, for all $v$ in $V$
$T(v)=\alpha w$, for $\alpha \in \mathbb{R}$.
I know that would be direct if i assume a orthogonal basis of $V$ but, at principle I would like to prove it for an ordinary basis $\{v_{1},...,v_{n}\}$.
Acctually, I was trying to put $w=v_{1}$ and then let the others vector become a basis to the Kernel of $T$, but i get no results. How can i use the inner producto information with an ordinary basis??