As Per Manne and EuYu commented, the positive definiteness only applies to symmetric matrices. Suppose $A$ is a symmetric square-matrix below. You can see here a basic example where it works.
Definitions for positively-definitess
determinant-of-all-squares -check: $det(A_{i,i})>0$ where $A$ is a symmetric quare-matrix and $i\leq n$ where $n$ is the degree of the matrix $A$
square-form -check: $\bar x^t A \bar x>0$ where $\bar x$ is inside an unit ball, defined in Finnish here where the unit-ball -definition here (pages 866-867 in the book here)
eigen-value -check: all eigen-values must be positive $\lambda_i>0 \forall i\in\mathbb N$
Generalization from Real numbers to Complex numbers
Suppose $z \in\mathbb C^n$. If $z^t M z \in\mathbb C^m$ with some $m$ and even with Hermitian $M$, then not positively definite, by this example here but please note that some authors use the notation $\mathbb R(z^tMz)>0$ where $z\in\mathbb C$ for positively definiteness. In other words, if the square -form is complex, it is not positively definite. You must get a scalar of the complex square form, otherwise the inequality undefined so not positively definite.
The latter Wikipedia -article mentions a relaxation of symmetrism with the matrices in complex case. This weaker form of positively-definiteness requires only that $z^t M z>0$ in which case the other definitions of positively-definiteness do not match, getting your ambiguous results. You can try this easily for example with your complex eigen-values: if you assume that $\mathbb R(z^tMz)>0$ only required for the definition, then you get Yes/No/Yes so not matching.
So your teacher is considering the strong form of positively-definiteness in which the definiteness have the same conclusion. In the weaker form, your "conflict" is correct: you both are correct depending on the definition of positively-definiteness, please, note that $a_{ij}=a^*_{ji}$ where $.^*$ is the complex conjuagate (needed in the Hermitian -definition, generalization of the term symmetric to complex case).