I'm having some confusion with the definition of the singular value of a matrix. As per Wikipedia:
In mathematics, in particular functional analysis, the singular values, or s-numbers of a compact operator $T : X \to Y$ acting between Hilbert spaces $X$ and $Y$, are the square roots of non-negative eigenvalues of the self-adjoint operator $T^*T$ (where $T^*$ denotes the adjoint of $T$).
But we know that the adjoint of a matrix $T$ coincides with its complex-conjugate transpose matrix $T^\dagger$. We also know that the eigenvalues of $T^\dagger$ as precisely complex conjugates of the eigenvalues of $T$. Moreover, we know that the eigenvalues of $T^\dagger T$ are non-negative as described here.
So if all eigenvalues of $T^\dagger T$ are already non-negative, then why does Wikipedia bother to consider square roots of only non-negative eigenvalues of $T^*T$? Aren't all eigenvalues of $T^*T$ non-negative in general, or am I missing something? (Here's a related question: Proving that the eigenvalues of $T^*T$ is non-negative)
All you said are right. Wikipedia is redundant but not wrong. The proof that the eigenvalues of $T^*T$ are all nonnegative should be more straightforward without picking basis or establishing the matrix form of $T^*$. If $\lambda$ is an eigenvalue with eigenvector $v$ of $T^*T$, then $$(T^*Tv, v)_X = (Tv, Tv)_Y\ge 0, \text{ also }(T^*Tv, v)_X = \lambda (v, v)_X$$ hence $$\lambda = \frac{(Tv, Tv)_Y}{(v,v)_X}\ge 0$$