Suppose that $V$ is an arbitrary vector space and that $T \in L(V)$ is invertible and admits an adjoint.
Is it true that, in this case, $T^*$ is invertible?
It is well-known that this is true in finite dimensional vector spaces, because, in those spaces, a transformation is invertible iff its injective, and $\ker(T^*) = \operatorname{Im}(T)^\perp = 0$.
But, in arbitrary ones, the equivalency falls short, and also, we can only say that $\operatorname{Im}(T^*) \subset \ker(T)^\perp$ - there are counter-examples to the other inclusion in arbitrary vector spaces.
It is easy to prove that if this is the case, then the adjoint's inverse will be an adjoint for the inverse of $T$ ($(T^*)^{-1} = (T^{-1})^*$), but I couldn't really prove existence before doing the algebraic manipulations...
In fact, perhaps it would be easier to prove that $T^{-1}$ admits an adjoint, since the result could also follow from there, but I also couldn't really move in that direction.
Does anyone have any insights as per how to proceed? Or perhaps a counter-example in infinite dimensions?
Thanks in advance!
Given vector spaces $V, W$ over a field $\mathbb{F}$, the adjoint of a linear map $T \colon V \to W$, is the linear map $T' \colon W' \to V'$ (here $V'$ denotes the space of all linear maps from $V$ to $\mathbb{F}$). defined by $\langle v, T'\omega\rangle := \langle Tv, \omega\rangle$, where for linear functional $b$, vector $a$, $\langle a, b \rangle = b(a)$. It follows easily that if $X$ is a vector space over $\mathbb{F}$ and $S \colon W \to X$ is linear, then $(ST)' = T'S'$.
Thus if $T \colon V \to V$ is an invertible linear map, then $T'(T^{-1})' = (T^{-1}T)' = I' = I_{V'}$ and $(T^{-1})'T' = (TT^{-1})' = I' = I_{V'}$. Thus $(T')^{-1} = (T^{-1})'$.
But it seems like you are asking about the adjoint when $V$ is an inner product space over $\mathbb{F} = \mathbb{R}$ or $\mathbb{C}$. If $V, W$ are inner product spaces over $\mathbb{F}$, then a linear map $T \colon V \to W$ has an adjoint $T^* \colon W \to V$ means that $(v, T^*w) = (Tv, w)$ for all $v \in V, w \in W$, where $(\cdot, \cdot)$ denotes inner product. In this case, I don't think all linear maps have adjoints. But if $V$ and $W$ are Hilbert spaces and $T \colon V \to W$ is a continuous linear map, then by the Riesz representation theorem, $T$ has an adjoint. Moreover, if $T$ is invertible, then by https://en.m.wikipedia.org/wiki/Bounded_inverse_theorem, $T^{-1}$ is continuous and therefore has an adjoint.