# Using singular value decomposition to assess approximate non-trivial solutions to A x = 0

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.

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, $$\| A v_i \| = \sigma_i \| u_i \| = \sigma_i.$$
• But $sqrt(0.0002952607) \neq \sigma_4$? Sep 7, 2020 at 1:52