This is a short note to self: Let $A$ be a symmetric positive semidefinite matrix with one-dimensional kernel spanned by $v$. How to solve $Ax=b$ (if you know that $b$ is in the range of $A$)? Just typing

`x = A\b`

should give a warning in a reasonable software (but also should produce some correct result, if it returns anything at all).

If you don’t want that warning and also want to get the solution that is orthogonal to the kernel you should do

`x = (A+v*v')\b.`

Note that $A + vv^T$ has full rank (and $v$ is still an eigenvector, but now for the eigenvalue $\|v\|^2$).

Surely, the solution of $Ax=b$ which is orthogonal to the kernel of $A$  also solves this $(A+vv^T)x = b$ since $(A+vv^T)x = Ax + vv^Tx = Ax = b$. Conversely, if $x$ solves $(A + vv^T)x = b$, then taking the inner product with $v$ gives $(Ax)^Tv + (v^Tx)^2 = b^Tv$ and since $b^Tv = 0$ and $(Ax)^T v = x^TAv = 0$ it follows that $v^T x = 0$ which shows that both $Ax=b$ and that $x$ is orthogonal to the kernel.

Also, if you want the solution which is orthogonal to some $z$ (and not to the kernel of $A$) you can solve $(A + zz^T)x=b$. By taking the inner product with $v$, you get that $v^T z\, x^T z=0$ and you get $x\bot z$ as soon as $v^Tz\neq 0$.