Given an $m \times n$ matrix $A$ and a vector $x\in\mathbb R^n$, with $m<n$, what's an efficient way of computing the projection of $x$ onto the kernel of $A$?


1 Answer 1


One part of the fundamental theorem of linear algebra is that the kernel/nullspace of $\mathbf A$ is orthogonal to the range of $\mathbf A^T$. By applying the $\mathbf Q \mathbf R$ decomposition to $\mathbf A^T$, you can generate the orthogonal projector $\mathbf P = \mathbf I - \mathbf Q \mathbf Q^T$. The vector $\mathbf P \mathbf x$ is what you're looking for. A brief matlab demo follows:

clear all
close all

% Form random A and x.
m = 23;
n = 39;
A = rand(m,n);
x = rand(n,1);

% Find Q = span(A')
[Q,~] = qr(A',0);

% Decompose x = Qx + Px
Qx = Q*(Q'*x);
Px = x-Qx;
norm_Px = norm(Px)
norm_Qx = norm(Qx)
error_x = norm(x-Px-Qx)

% Verify Px is in nullspace of A.
error_APx = norm(A*Px)

If $\mathbf A$ is too large but has exploitable structure (sparsity? some kind of H-matrix like rank-deficiency?), you might be better off using using randomized sampling / Krylov ideas, instead of dense $\mathbf Q \mathbf R$ decomposition.

  • 2
    $\begingroup$ Good answer! Only one remark: this method only works when that $A$ has full row rank. Otherwise one needs to use the first factor of a rank-revealing decomposition (QRP or SVD). $\endgroup$ Commented Jul 9, 2020 at 6:50
  • $\begingroup$ @FedericoPoloni Can you expand? I'm not sure what you mean by "rank-revealing decomposition" $\endgroup$
    – a06e
    Commented Sep 9, 2021 at 22:42
  • $\begingroup$ @becko Some decompositions give you numerically stable information on the rank of a matrix; among them QRP and SVD. QR is not one of them. There is a discussion of this in the Golub-Van Loan book, for instance. $\endgroup$ Commented Sep 10, 2021 at 7:17
  • $\begingroup$ @FedericoPoloni But why is this relevant in this context? Why does the above answer fail when $A$ is not full rank? $\endgroup$
    – a06e
    Commented Sep 10, 2021 at 7:19
  • 1
    $\begingroup$ @becko It's easy to construct examples where it fails; for instance taking A = [1 0; 0 0] and a QR decomposition with $Q = I$; then $P=0$ is wrong. $\endgroup$ Commented Sep 10, 2021 at 9:19

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