Is this operator $A = \pmatrix{1&1\\0&1}$ self-adjoint? Is this operator
$$A = \pmatrix{1&1\\0&1}$$
self-adjoint? I think not, because
$$\pmatrix{1&1\\0&1}^T\neq A$$
where $T$ is the transposition of the matrix. What do you all think?
 A: Assuming this is an operator on $\mathbb{R}^2$ with the standard inner product $\left< x, y \right> = x \cdot y$, then yes, this is the correct reasoning.
In general, you need to specify an inner product to know if an operator is self-adjoint, because self-adjointness means $\left< x, A y \right> = \left< A x, y \right>$.
In this case, notice that $\left< x, A y \right> = x^T A y$ whilst $\left< A x, y \right> = (Ax)^T y = x^T A^T y$. Consequently one needs $A = A^T$.
A: No, the operator is not self adjoint.  Your reasoning is correct.
Note that the definition of the adjoint of an operator is not simply that the adjoint of $A$ is $A^T$.  However, in the context of real matrices taken with respect to the standard dot product or with respect to the standard dual basis, this is what "adjoint" comes to mean.  An operator is "self-adjoint" if it is equal to its adjoint, which for real matrices means that $A = A^T$.
A: In fact, for any scalar product, $A$ is never self-adjoint. Let $f$ be a non-degenerate symmetric bilinear form on $\mathbb{R}^2$ and $M=\begin{pmatrix}a&b\\b&c\end{pmatrix}$ be its matrix in the canonical basis. The adjoint $\tilde{A}$ of $A$ is defined by $f(x,\tilde{A}y)=f(Ax,y)$, that is $M\tilde{A}=A^TM$. Thus $A$ is self-adjoint iff $MA=A^TM$. We obtain the following conditions on $M$: $a=0,b\not= 0$, and $f$ cannot be positive.
EDIT. We prove the PROPOSITION (valid for any $n$): Let $A\in M_n(\mathbb{R})$. There is a scalar product $f$ s.t. $A$ is self-adjoint with respect to $f$ $\iff$ $A$ is diagonalizable over $\mathbb{R}$.
PROOF. ($\Rightarrow$) According to Andreas's comment below, if $A$ is $M$-self-adjoint, then $A$ can be diagonalized with a $M$-orthogonal matrix $U$ (it satisfies $f(x,y)=f(Ux,Uy)$ that is $M=U^TMU$). 
($\Leftarrow$) There is an invertible $P$ s.t. $A=PDP^{-1}$ where $D$ is diagonal. Let $\Delta$ be a positive diagonal matrix and $M=P^{-T}\Delta P^{-1}$; note that $M$ is symmetric $>0$. Clearly $MA=A^TM$ and we are done.
