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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.

$$N(A+B)⊂N(A)$$ $$N(A+B)⊃N(A)$$

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  • $\begingroup$ 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 $N(A) \cap N(B) \subset N(A+B)$. $\endgroup$ – TrostAft Dec 10 '18 at 7:49
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First one false when $A=-B=I$. the second one is false when $A=0$ and $B=I$.

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  • $\begingroup$ For the first case, I think $A =I$ . $\endgroup$ – AnyAD Dec 10 '18 at 7:51
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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 $(A+B)x=0\implies Ax=0$.

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