Null Space of Sum of Two Matrices is a subset or supset of null space of one

Could anyone explain how either of these can be proven? I don't see how either of these statements by themselves can be true, much less how to prove them.

$$N(A+B)⊂N(A)$$ $$N(A+B)⊃N(A)$$

• Suppose $x \in N(A+B)$. This implies $(A+B)x = 0 \implies Ax + Bx = 0$. Is it necessarily true that $Ax = 0$? What about the other way around, if we have $Ax = 0$, then must it be true that $(A+B)x$? In fact I don't believe either of your statements, but I do buy that $N(A) \cap N(B) \subset N(A+B)$. – TrostAft Dec 10 '18 at 7:49

First one false when $$A=-B=I$$. the second one is false when $$A=0$$ and $$B=I$$.
• For the first case, I think $A =I$ . – AnyAD Dec 10 '18 at 7:51
They will/may be true for special choice/s of $$A,B$$ but they are not true in general as shown in the other answer.
For the case that this is true, you'd prove the first statement fir example by taking $$x\in N (A+B)$$ and showing that this implies also that $$x\in N(A)$$. That is $$(A+B)x=0\implies Ax=0$$.