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$$ S_1 = \{B \in \Bbb R^{2\times2} \mid BA = 0\} $$

I know that to check if a set is a subspace of a vector space, I need to determine if it is closed under addition and subtraction. I honestly have no idea how to do this for this set.

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I changed $2x2$ to $2\times2$. – Michael Hardy Oct 2 '12 at 15:59
up vote 3 down vote accepted

You have to check that if $B,C$ in $S_1$, and $\alpha\in\Bbb R$, then $B+C\in S_1$ and $\alpha B\in S_1$. I’ll do the second one; the basic ideas are the same in both cases.

To show that $\alpha B\in S_1$, you must show that it satisfies the condition for membership in $S_1$: you must show that $(\alpha B)A=0$. By the properties of matrix arithmetic you know that $(\alpha B)A=\alpha(BA)$. By hypothesis $B\in S_1$, so $BA=0$. Thus, $(\alpha B)A=\alpha(BA)=\alpha 0=0$.

You'll use the same basic approach to show that $B+C\in S_1$, though the specific properties of matrices that you’ll use are different.

Oh, I almost forgot: you also have to check that $S_1\ne\varnothing$, which you can do by finding just one matrix that’s definitely in $S_1$. Since you don’t know just what matrix $A$ is, that might sound hard, but it really isn’t: there’s one $2\times 2$ matrix whose product with $A$ is certainly $0$.

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