SVD and homogeneous equation

Question

Suppose a $m \times n$ matrix $A$, and column vector $h$.
($A$'s rank is equal or smaller then $n$(=$h$'s length).)

If, $$ Ah=0 $$

then $h$ can be the last column of $V$ where $A = UDV^T $.

($UDV^T$ is taken by SVD)

How can I prove this?

Answer 1

If $v_i$ is the $i$th right singular vector, $\sigma_i$ is the $i$th singular value, $u_i$ is the $i$th left singular vector, and $e_i$ is the $i$th standard basis vector, then

$$Av_i=U D V^T v_i = U D e_i = U (\sigma_i e_i)=\sigma_i u_i.$$

The first step follows by the fact that $V$ is orthogonal, the others follow from $Be_i=b_i$ for any matrix $B$.

Can you conclude?

Answer 2

Homogenous equations $Ax=0$ only have two possible solutions, the trivial solution $(x_1=x_2=...=x_n=0)$ or infinitely many solutions depending on the rank of the matrix. If the matrix is rank-deficient, there are singular values that are zero. This means that $σ_i$ is zero and thus Ian's answer completes to $Av_i = 0$, which makes $v_i$ a solution to the equation. In a full rank case, all singular values $σ_i$ are non-zero and thus there only exists the trivial solution.