Suppose $A$ and $P$ are two nxn matrices. Is $ N(PA) \subseteq N(A)$ ? How would you prove that, or disprove that?
I tried to prove this and realized that if I do not know if $A$ and $P$ are nonsingular, then I cannot use the simple proof utilizing $ P^{-1} $.
What I reasoned was that I have two cases for $ A\vec{x}=0 $, where I have $ \vec{x} = 0 $, and $ \vec{x} \neq 0 $.
Where $ \vec{x} = 0 $, since the only solution here is the trivial solution, we know that $PA$ must be nonsingular, hence both $A$ and $P$ are nonsingular. Letting $ \vec{x}\in N(PA) $. Then $ PA\vec{x} = 0$. We can show that $ P^{-1}PA\vec{x} = P^{-1}0 \Rightarrow A\vec{x} = 0$. Therefore, $ \vec{x}\in N(A) $.
Where $ \vec{x} \neq 0 $, I reasoned that either $A$ and $P$ are nonsingular, or that $A$ or $P$ is singular.
If $A$ and $P$ are nonsingular, then the proof that $ N(PA) \subseteq N(A) $ is pretty straightforward like the last one. But if they are not both nonsingular, then we do not necessarily know if the rank is preserved under the product of $A$ and $P$. If the $ rank(PA) \neq rank(A) $, then they cannot be the same subspace.
2026-03-04 17:50:55.1772646655