Why are singular values always non-negative?

I'm assuming that the matrix $A$ has real entries, or else you should be considering $A^*A$ instead.

If $A$ has real entries then $A^TA$ is positive semidefinite, since $$ \langle A^TAv,v\rangle=\langle Av,Av\rangle\geq 0$$ for all $v$. Therefore the eigenvalues of $A^TA$ are non-negative.

Assume that $A$ is real for simplicity. The set of (orthogonal,diagonal,orthognal) matrices $(U, \Sigma, V)$ such that $A = U \Sigma V^T$ is not unique. Indeed, if $A = U \Sigma V^T$ then also

$$ A = (-U)(-\Sigma)V^T = U (-\Sigma)(-V^T) = (UD_1)(D_1 \Sigma D_2)(V D_2)^T$$ for any diagonal matrices $D_1$ and $D_2$ with only $1$ or $-1$ on the diagonal. Therefore, the positivity of the singular values is purely conventional.

Suppose $T \in \mathcal{L}(V)$, i.e., $T$ is a linear operator on the vector space $V$. Then the singular values of $T$ are the eigenvalues of the positive operator $\sqrt{T^* \; T}$. The eigenvalues of a positive operator are non-negative.

• Why is $\sqrt{T^* \; T}$ a positive operator? Consider $S = T^* \; T$. Then $S^* = (T^* \; T)^* = T^*\;(T^*)^* = T^*\;T=S$, and hence $S$ is self-adjoint. Also, $\langle Sv, v \rangle = \langle T^*\,T v, v \rangle = \langle Tv, Tv \rangle \geq 0$ for every $v \in V$. Hence $S$ is positive. Now every positive operator has a unique positive square root, which, for $S$, I am denoting with $\sqrt{T^* \; T}$.
• Why are the eigenvalues of a positive operator non-negative? If $S$ is a positive operator, then $ 0 \leq \langle Sv, v \rangle = \langle \lambda v, v \rangle = \lambda \langle v, v \rangle$, and thus $\lambda$ is non-negative.