I learned about simple value decompositions from my online Linear Algebra class today, and one thing I learned in particular was that for any real $m \times n$ matrix $A$, $A^TA$ has only nonnegative eigenvalues. I’m wondering how this might extend to imaginary/complex numbers — is it possible for $A^T A$ to have complex eigenvalues?
EDIT: To clarify, I meant to ask if for a matrix $A$ with all real entries, does $A^T A$ have any complex eigenvalues in the form $a+bi$ where $b\neq 0$?
Yes, for a complex matrix $A$, it is possible for $A^TA$ to have non-real complex eigenvalues. An example given in the comments is the $1\times 1$ matrix $[1+i]$. (If $A$ is a complex matrix whose entries all just happen to be real, then it is no different from the real case. And in the real case, $A^TA$ must have non-negative real eigenvalues.)
However, if we in addition to transposing $A$ also conjugate every element in it, we get what is known as the "conjugate transpose" of $A$. It is usually denoted $A^*$, but you might also come across things like $A^H$ (for "Hermitian conjugate") or $A^\dagger$. And for any complex matrix $A$, we have that $A^*A$ only has nonnegative real eigenvalues. (If $A$ is a complex matrix whose entries all just happen to be real, then $A^* = A^T$, and thus $A^TA$ has only non-negative real eigenvalues, just as in the real case.)
In complex linear algebra, we rarely transpose, and almost always conjugate transpose. It is the conjugate transpose that carries over most of the algebraic properties that the transpose has in real linear algebra.