Given an $N \times 4$ matrix $A$ $(N \geq 4),$ I can find an approximate (non-trivial) solution to the homogeneous equation $A x = 0$ using singular value decomposition (SVD) assuming that one of the singular values is sufficiently small. Say the SVD yields $A = U \Sigma V^T$ and the smallest singular value is $\sigma,$ then the corresponding column of $V$ yields the approximate solution $\hat{x}.$ Can I quantify how good this solution is based on $\sigma$? Perhaps $\sigma$ tells me something about the magnitude of $\| A \hat{x} \|$?
For example, the matrix
$$
A = \left[\begin{array}{cccc}
2 & 2 & 2 & 2.3 \\
4 & 4 & 4 & 4 \\
-1 & -1 & -1 & -1 \\
5 & 5.1 & 5 &5 \\
3.1 & 3 & 3 &3
\end{array}
\right]
$$
has singular values
$\sigma_1 = 14.9275,$ $\sigma_2 = 0.260285,$
$\sigma_3 = 0.0986021,$ $\sigma_4 = 0.0322844.$ The last column of
$V$ is $\hat{x} = [-0.313176,\ -0.457528,\ 0.830508,\ -0.0533385]^T$ which corresponds
to $\sigma_4.$ We see this solution is close:
$$
A \hat{x} = \left[\begin{array}{c}
-0.00307055 \\
0.025862 \\
-0.0064655 \\
-0.0134253 \\
-0.0119211
\end{array}\right] \approx \mathbf{0}
$$
(check here) and $\| A \hat{x} \| = 0.0322866.$
We know the solution is normalized ($\| \hat{x} \|^2 = 1$), but can we quantify
the error in the solution based on $\sigma_4?$
Edit: fixed arithmetic errors which match the answer.
Best Answer
Indeed, $\| A \hat x \| = \sigma$.
Note that $A = U \Sigma V^T \implies AV = U \Sigma$. Column by column, we see that $A v_i = \sigma_i u_i$, where $u_i$ and $v_i$ are the $i$th columns of $U$ and $V$, respectively, and $\sigma_i$ is the $i$th singular value of $A$.
So, $$ \| A v_i \| = \sigma_i \| u_i \| = \sigma_i. $$