Let $H$ be a Hilbert space with $\{e_n\}$ an orthonormal basis. Then, which is false for $T:H\to H$:
a)$T(e_n)\neq e_1\forall n$
b)$T(e_n)=e_{n+1}\forall n$
c)$T(e_n)=\sqrt{\frac{n+1}{n}e_n}\forall n$
d)$T(e_n)=e_{n-1}\forall n$
I think a) is false, because the linear operator can just map any element of the basis to any other element? Am I right in that? What about other options. Thanks beforehand.
consider the space of square-summable sequences $ℓ^2 = \{ (u_n)_{n\geq 1} \subset \mathbb{C}, \; \sum_{n \geq 1}|u_n|^2 < \infty \}$
it is a Hilbert space with respect to the inner product $(u_n,v_n) = \sum_{n \geq 1} u_n \overline{v_n}$ and it's got an orthonormal sequence $\{e_n\}$ defined as follows : $$e_n = (0,0,0, \cdots,1,0,0,\cdots,0,\cdots)$$
the $1$ is in the $n$-th slot
consider the operator
$\begin{align}A_n : &ℓ^2 \to ℓ^2 \\ &e_k \mapsto e_1 \text{ if } k = n, 0 \text{ if } k \neq n\end{align}$
if $X = (X_1,X_2,\cdots,X_n,\cdots) \in ℓ^2$ then clearly $A_nX = X_ne_1 = (X_n,0,0,\cdots,0,\cdots)$
$A_n$ is obviously linear.
also $\| A_nX \|^2 = \| X_ne_1 \|^2 = \sum_{n \geq 1}|X_ne_1|^2 = |X_n|^2 \leq \|X\|^2$
this proves the boundness.
and it's easy to notice that $A_ne_n = e_1$
a) is false then