= Solution
Using the signs chosen above, explicit multiplication gives
$$
\boxed{A\mathbf v_1=\frac1{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}=\mathbf u_1,\qquad
A\mathbf v_2=\begin{pmatrix}1\\-1\end{pmatrix}=\sqrt2\,\mathbf u_2,\qquad
A\mathbf v_3=\mathbf0.}
$$
Thus the nonzero cases obey $A\mathbf v_m=\sqrt{\sigma_m}\,\mathbf u_m$. This relates the <right singular vectors> and <left singular vectors> in a <singular value decomposition>. For example, if $U=(\mathbf u_1\ \mathbf u_2)$ and $V=(\mathbf v_1\ \mathbf v_2\ \mathbf v_3)$, then $A=U\Sigma V^{\mathsf T}$ with $\Sigma=\begin{pmatrix}1&0&0\\0&\sqrt2&0\end{pmatrix}$.
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