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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.

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4 Answers

up vote 1 down vote accepted

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 $\ker (A+B)$.

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Thank you very much sir. –  srijan Apr 5 '13 at 8:24
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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!

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Hint: Let $A$ be any non-zero matrix and let $B$ be the zero matrix (and vice versa).

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Thank you very much for your response. I am taking both $A$ and $B$ as a non zero matrix. –  srijan Apr 5 '13 at 7:19
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That's not what I recommended in my answer. Either let $A$ be non-zero and let $B$ be zero, or let $A$ be zero and let $B$ be non-zero. What happens in either case? (What is the null space of the zero matrix?) –  Zev Chonoles Apr 5 '13 at 7:20
    
@Zevv If B is null matrix then every vector in the corresponding space will be null space of B. Perhaps I am not able to get your hint :(. –  srijan Apr 5 '13 at 7:24
    
That's correct; every vector is in the null space of the zero matrix. Now consider this: is it possible for every vector to be in the null space of a non-zero matrix? –  Zev Chonoles Apr 5 '13 at 7:25
    
Thanks for your help sir. –  srijan Apr 5 '13 at 8:24
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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 $v \in N(A+B)$. $\blacksquare$

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farin Thanks for your reply. Could you give me proof of your assertion? It would be very helpful for me. Thanks –  srijan Apr 5 '13 at 8:23
    
@ Lord farin Thanks for your answer. :) –  srijan Apr 5 '13 at 8:46
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