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?

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 $A = A^T$.
