I am reading that a linear operator $A$ can be thought of as a (1,1) tensor [where (r,s) corresponds to r vectors and s dual vectors]. This can be done by saying $$A(v,f) \equiv f(Av)$$ where v is a vector and f is a dual vector.

I assume the $Av$ in the argument on the rhs corresponds to the linear operator $A$ acting on the vector $v$. But this is where I get mixed up. What is the dual vector for $Av$ if $A$ is a (1,1) tensor? And isn't a tensor supposed to be a multilinear map that results in a number? If that is the case, why does $Av$ presumably result in a vector that goes in the argument of $f$ on the rhs? What is $Av$ here?

I am reading that a linear operator $A$ can be thought of as a (1,1) tensor [where $(r,s)$ corresponds to $r$ vectors and $s$ dual vectors]. This can be done by saying $$A(v,f) \equiv f(Av)$$ where $v$ is a vector and $f$ is a dual vector.

I assume the $Av$ in the argument on the rhs corresponds to the linear operator $A$ acting on the vector $v$. But this is where I get mixed up. What is the dual vector for $Av$ if $A$ is a (1,1) tensor? And isn't a tensor supposed to be a multilinear map that results in a number? If that is the case, why does $Av$ presumably result in a vector that goes in the argument of $f$ on the rhs? What is $Av$ here?