How does a linear operator act on a bra? I'm studying QM from Shankar. He introduces linear operators and says that an operator is an instruction for transforming one ket into another. But then a few lines below he says operators can also act on bras.
So does the complete specification of an operator include its action on bras? Or does its action on the kets determine its action on the bras? 
 A: *

*Using Shankar's notation from his eq. (1.6.14), an operator acting on a bra$^1$
$$\langle \phi | ~=~ \langle \psi |\hat{A} $$
is equivalent to the Hermitian adjoint operator acting the corresponding ket
$$|\phi \rangle   ~=~ \hat{A}^{\dagger}|\psi \rangle .  $$

*The Hermitian adjoint operator fulfills the relation
$$  \langle w | \hat{A}v \rangle~=~\langle \hat{A}^{\dagger}w | v \rangle.  $$
--
$^{1}$We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.
A: There is a standard definition for the action of a linear operator on a bra, although many authors (including Shankar) sloppily skip defining it. $\langle \phi| A$ is defined to be the linear functional such that $\big( \langle \phi| A \big) |\psi\rangle \equiv \langle \phi| \big(A |\psi\rangle\big)$. In other words, you first act the linear operator on the ket that's getting "eaten" and then act the bra $\langle \phi |$ on the output of that.
Similarly, we can define the action of the adjoint of a linear operator on a ket. $$A^\dagger | \psi \rangle$$ is defined by how an arbitrary bra $\langle \phi |$ acts on it:
$$\langle \phi | \big( A^\dagger | \psi \rangle \big) := \big( \langle \phi | A^\dagger \big) | \psi \rangle = \langle A \phi | \psi \rangle.$$ Note that unlike the former case, this requires an inner product structure so that we can define the adjoint.
The inner product of a ket $A^\dagger | \psi \rangle$ with all possible kets $|\phi\rangle$ completely specifies the ket, because the positive-definiteness of the inner product implies that it's non-degenerate. It's not often appreciated that the adjoint of an operator only has a unique action on kets because of the positive-definiteness requirement on the inner product.
A: The actual statement holds in a much more general setting. 
Let $\Phi:V\to V$ be any mapping, and let $f:V\to K$ be a function on $V$. Then we get a new function $\Phi^\ast f:V\to K$ defined by $v\mapsto f(\Phi(v))$.
In your case, $V$ is the space of kets, and $\Phi$ is a linear operator on it. A linear map $f:V\to\mathbb C$ is a bra. (Let's stay in the finite dimensional case to not have to worry about continuity and so.) Since $\Phi$ is linear, it is not hard to see that if $f$ is linear, then so is $\Phi^\ast f$. That is all there really is about how $\Phi$ acts on bra's. 
If you take this further, then you see how the matrix of $\Phi$ on kets transforms when expressed in the dual basis of bra's. This is elaborated in the other answers.
A: If $A:V\to V$ is a linear operator on a vector space $V$ with basis $|m\rangle$,  Dirac defined the action $\langle n|A$ of $A$ on the bra vector $\langle n|$ (an element of the dual space $V^*$)as a new bra vector  that is evaluated  on a ket $|m\rangle$ to give the same answer as the evaluation of $\langle n|$ on $A|m\rangle$. In other words he set
$$
(\langle n|A)|m\rangle = \langle n|(A|m\rangle).
$$
Since it does not matter where one puts the brackets Dirac just writes 
$$
\langle n|A|m\rangle
$$
for both things. Note that the dagger operation $\dagger: |m\rangle\mapsto \langle m|= (|m\rangle)^\dagger$ that provides an antilinear map between $V$ and $V^*$ (and so defines the inner product on $V$) is not needed here. So $A$ acting on bras has nothing to do with the hermitian conjugate $A^\dagger$ which needs the inner product. As $A$ acting to the left on bra vectors is really an operator $V^*\to V^*$  rather than an operator $V\to V$ it usual in linear algebra  to regard it as different operator $A^*:V^*\to V^*$ that is called the "conjugate," or the "transpose."  The latter name is probably best as no complex conjugation is involved, and in the dual basis $A^*$ is representated by the transpose (but not the complex conjugate transpose)   of the matrix representing $A$. 
Note added: I just saw Qmechanic's answer. What he says: $A|\phi\rangle = |\chi\rangle \Leftrightarrow \langle \phi|A^\dagger =\langle \chi|$ is quite correct, but you really don't need to use the dagger to define left action of $A$.
