How to find a matrix whose range is the null space of another matrix?

Question

One solution for eliminating the equality constraint from optimisation problems is employing a typical matrix $F$ which its range space is the null space of the matrix used in equality constraint as follows:

The original problem, which is a convex problem: $$ min f_0(x) \\ s.t. f_i(x) \le 0, i =1,...,m \\ Ax = b $$

and the eliminated equality constraint problem:

$$ min f_0(Fz+x_0) \\ s.t. f_i(Fz+x_0) \le 0, i=1,...,m $$

What I've done is that after searching alot I found that I should use the SVD decomposition of the matrix A in equality which is $A=U \Sigma V^T$ and multiply it with $Fz$, and this should be equal to zero:

$$ (A)Fz=(U \Sigma V^T)Fz=0 $$

but I cannot find a property which satisfies the above equation. Does anyone have any opinion?

I have to point out I've seen Eliminating equality constains, but it's just an explanation of employing this technique, and it does not find the matrix $F$.

Answer 1

If we have the SVD of $A$, then the following approach is possible. Partition the matrices $U,\Sigma,V$ so that $$ \Sigma = \pmatrix{\Sigma_0 & 0\\0 & 0}, \quad V = \pmatrix{V_1 & V_2} $$ where $\Sigma_0$ is $r \times r$ with non-zero diagonal entries, and $V_1$ has $r$ columns. The nullspace of $A$ is spanned by the columns of $V_2$. Thus, $F= V_2$ is a matrix whose range is the nullspace of $A$.

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In response to the comment:

Suppose that $A$ has size $m \times n$. Recall that $U,V$ are orthogonal (and therefore invertible) matrices. So, we have $$ Az = 0 \iff U(\Sigma V^T)z = 0 \iff \Sigma (V^Tz) = 0. $$ We see that $z$ is in the nullspace of $A$ if and only if $V^Tz$ is in the nullspace of $\Sigma$. However, if we break the column-vector $w$ into $w = w_1,w_2$ (with $w_1$ of length $r$), then we find (with block-matrix multiplication) $$ \Sigma w = \pmatrix{\Sigma_0 & 0\\ 0&0} \pmatrix{w_1\\w_2} = \pmatrix{\Sigma_0 w_1\\ 0}. $$ In other words, the nullspace of $\Sigma_0$ is spanned by the columns $e_{r+1},\dots,e_n$ (where $e_k$ is the $k$th standard basis vector of $\Bbb R^n$, i.e. the $k$th column of the $n \times n$ identity matrix). This means that the vectors $z_{r+1},\dots,z_n$ that solve $V^Tz_k = e_k$ form a basis of the nullspace. With that, we see that $$ V^Tz_k = e_k \implies z_k = Ve_k, $$ which is to say that $z_k$ is the $k$th column of $V$. So, the columns $V e_{r+1},\dots,V e_n$ of $V$, i.e. the columns of $V_2$, form a basis of the nullspace.

Answer 2

I don't know anything about constraints, but supposing for a moment that you actually have a basis $b_1, \ldots, b_k$ for the nullspace of $A$, i.e., that $$ Ab_i = 0 $$ for each $i$, and $Ax = 0$ implies there are coefficients $c_i$ such that $$ x = c_1 b_1 + \ldots c_k b_k, $$ then creating such a matrix is straightforward: you use the vectors $b_i$ as the columns of a matrix $B$.

This in fact works (in the sense that it literally answers your question) even if the $b_i$ are only a spanning set rather than a basis (i.e., they need not be linearly independent), although my guess is that linear independence of the $b_i$s is probably a nice way to make your problem smaller.

Let's see an example. Let $$ A = \pmatrix{1 & 1 & 0 \\ 0 & 1 & 2 \\ 2 & 1 & -2}. $$ Then $A$ has rank $2$, so its nullity is $1$, and in fact $$ b_1 = \pmatrix{2 \\ -2 \\1} $$ spans the nullspace. When we make a matrix $B$ out of this single column, we get $$ B = \pmatrix{2 \\ -2 \\1} $$ whose image is, in fact, exactly the nullspace of $A$.