Null space of the sum of two matrices If we have two square matrices A and B then can we claim that $N(A + B)\supset N(A) $ and $N(A + B)\supset N(B)$,  respectively ? I have this doubt. Could anybody help me with this? I would be very much thankful to you.
 A: As the other answers have indicated, your assertions are not true. However, a moment's thought does reveal that we can salvage a partial result:
$$N(A+B) \supseteq N(A)\cap N(B)$$
which effectively rests on the fact that $0 + 0 = 0$ in any vector space.

As per request, a proof of the above. We are asserting that $v \in N(A) \cap N(B)$ implies $v \in N(A+B)$. Now suppose that the premise holds, i.e. $v \in N(A)\cap N(B)$:
$$\begin{align*}(A+B)v &= Av + Bv & & \text{by definition of $(A+B)v$} \\
&= 0 + 0 & & \text{since $v \in N(A)$ and $v \in N(B)$} \\ &= 0\end{align*}$$
hence $v \in N(A+B)$. $\blacksquare$
A: two systems (A+B)x=0 and Ax=0 do not have same answer always!
so $N(A + B)\supset N(A)
$ is not always correct!
A: Take $A=\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$, $B=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$. Then $\ker (A+B) = \{0\}$, $\ker A = \operatorname{sp} \{ e_1 \}$, $\ker B = \operatorname{sp} \{ e_2 \}$.
Clearly, neither $\ker A$ nor $\ker B$ are contained in $\ker (A+B)$.
A: Hint: Let $A$ be any non-zero matrix and let $B$ be the zero matrix (and vice versa).
