I'm working through Lang's Linear Algebra, and a problem I'm working on a problem that has a hypothesis I'm not sure of. Let $ A:V\to V $ by a symmetric linear map. Show that $Ker(A) = \{0\}\ $ if and only if no eigenvalue is zero. My idea is this:
Assume $Ker(A) = \{0\}$. Then $Av = \lambda v = 0$ if and only if $v=0$. So, if $\lambda = 0$, we have that $Aw = 0$ for any $w\in V$. But this contradicts the assumption that the kernel has one element. Hence, $\lambda = 0$. Conversely, suppose $\lambda \neq 0 $. Then if $Av = 0 = \lambda v$, we have that $v=0$, so that $Ker(A) = \{0\} $.
However, nothing in this proof relies on $A$ being symmetric. Did I make an error in my proof, or is $A$ being symmetric an extraneous condition?
I believe you meant $A: V \to V$, not $L: V \to V$. And yes, the symmetry condition seems redundant because in general, for any $\lambda \in F$, where $F$ is the field we are working with, we have \begin{align} \ker(A- \lambda I) \neq \{0\} & \iff \text{there is a non-zero $v \in V$ such that } A(v) = \lambda v \\ & \iff \text{$\lambda$ is an eigenvalue of $A$} \end{align}
So, by taking the contrapositive, and choosing $\lambda = 0$, we get the statement: