Uncountable linearly independent set of vectors implies the space is not separable? + A question on self adjoint operators Let $\mathcal H$ be a hilbert space and let $A\subset \mathcal H$ be an uncountable set of linearly independent vectors. Does this imply $\mathcal H$ is not separable?
If $A$ was a set of orthonormal (or orhtognal) vectors then $\mathcal H$ is indeed not seprable.
What about the more general case? I don't have any idea on how to proceed from here, but intuitively i would say that it doesn't imply anything, meaning $\mathcal H$ can be separable and it can be not-separable.
Another question regarding self-adjoint operators:
Suppose $T:\mathcal H \rightarrow \mathcal H$ is a (linear) self adjoint operator. We also assume $T^k \equiv 0$ for some $k\in\mathbb N$. What can we say about $T$?
If we further assume that $T$ is of finite rank, meaning $\dim \text{Im} T < \infty$ then we can reduce the problem to a linear algebra problem and conclude $T\equiv 0$
Can this be generalized?
 A: The answer to the first question is negative. The separable Hilbert space $l^2$ contains an uncountable linearly independent set. For example, any Hamel basis (over $\mathbb C$, or over $\mathbb R$ if you prefer real Hilbert spaces) is uncountable. For a more explicit example, recall first that $\mathbb N$ has an uncountable family of infinite subsets $A_r$, one for each real number $r$, such that the intersection of any two of them is finite. (Proof: Fix an enumeration of $\mathbb Q$ as $\{a_n:n\in\mathbb N\}$, choose for each real $r$ a sequence $S_r$ of distinct rationals converging to $r$, and set $A_r=\{n:a_n\in S_r\}$.)  Now define, for each real $r$, an element $x_r\in l^2$ by letting the $n$th component of $x_r$ be $1/n$ if $n\in A_r$ and zero if $n\notin A_r$.  It is easy to check that these $x_r$'s are linearly independent.
For the second question, notice that $T^k=0$ implies that the spectrum of $T$ is $\{0\}$, so its spectral radius is $0$, and therefore $T=0$.
For a more down-to-earth answer to the second question, notice that, if we have $T^{2n}=0$ for a self-adjoint $T$, then we also have $T^n=0$.  Proof: For every $x$,
$$
(T^n x,T^nx)=(x,T^{2n}x)=(x,0)=0 \text{, so } T^nx=0.
$$
So from $T^k=0$ we can get $T^{\lceil k/2\rceil}=0$.  Applying this repeatedly, we eventually get $T=0$.
