= Solution
In $A^{\mathsf T}A$, every row sums to one, so $(1,1,1)$ is an <eigenvector> with <eigenvalue> $1$. The antisymmetric direction $(1,0,-1)$ gives <eigenvalue> $2$. A direction orthogonal to both is $(1,-2,1)$, and direct multiplication shows that it lies in the <kernel of a linear map>. Normalizing all three gives the complete system
$$
\boxed{\begin{array}{c|c}
\sigma_1=1&\mathbf v_1=\dfrac1{\sqrt3}(1,1,1)^{\mathsf T}\\[3pt]
\sigma_2=2&\mathbf v_2=\dfrac1{\sqrt2}(1,0,-1)^{\mathsf T}\\[3pt]
\sigma_3=0&\mathbf v_3=\dfrac1{\sqrt6}(1,-2,1)^{\mathsf T}
\end{array}}
$$
They form an <orthonormal basis> of $\mathbb R^3$. The extra zero <eigenvalue> reflects the one-dimensional <kernel of a linear map> for this rank-two rectangular <matrix>. The question's symbols $\sigma_m$ denote the <eigenvalues> of $A^{\mathsf T}A$; the nonzero <singular values> of $A$ are their <square roots>, namely $1$ and $\sqrt2$.
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