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A and B are given $n\times$ m matrices If $null(A) \subset null(B)$ what conclusion can we draw about range Space of $A$ and $B$. Can we conclude that range space of B is contained in a range space of A?

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Not really. Compare matrices like $\begin{pmatrix}1 \\ 0 \end{pmatrix}$, $\begin{pmatrix}0 \\ 1 \end{pmatrix}$, ... –  Dylan Moreland May 8 '12 at 3:39
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You can only say that the rank of $B$ is less than or equal to the rank of $A$, which follows immediately from the rank-nullity theorem: $\mbox{dim }N(A) + \mbox{dim }R(A) = n$. –  user12014 May 8 '12 at 3:42
    
What if $null (A)\subseteq null (B)$ –  srijan May 8 '12 at 3:45
    
Doesn't matter. You still can't get the result. –  Arturo Magidin May 8 '12 at 3:45
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up vote 3 down vote accepted

No, you can't conclude that the range of $B$ is contained in the range of $A$. You can conclude that $\dim(\mathrm{range}(B))\leq\dim(\mathrm{range}(A))$, by the Rank-Nullity Theorem, but then you run into the exact same problem that we had with your previous question: just because $\dim(\mathrm{range}(B))\leq \dim(\mathrm{range}(A))$, you cannot conclude inclusion of subspaces except in two trivial cases: when $\dim(\mathrm{range}(B))=0$, and when $\dim(\mathrm{range}(A))=n$.

So the only situation in which $\mathrm{range}(B)\subseteq\mathrm{range}(A)$ would follow is when $n\leq m$ and $\mathrm{nullity}(A)=m-n$, or when $\mathrm{null}(B)=\mathbb{R}^m$; because in the first case we would have $\mathrm{range}(A)=\mathbb{R}^n$ and in the latter we would have $\mathrm{range}(B) = \{0\}$.

To see this, note that given any two subspaces $W$ and $W'$ of $\mathbb{R}^n$ of the same dimension, there always exists an $n\times n$ matrix $C$ that is invertible, and such that $C(W)=W'$. Suppose you were able to conclude that $\mathrm{range}(B)\subseteq\mathrm{range}(A)$, and that $\mathrm{range}(A)\neq\mathbb{R}^n$ and $\mathrm{range}(B)\neq\{0\}$. Then there is a subspace $W$ of $\mathbb{R}^n$ of the same dimension of $\mathrm{range}(B)$ but that is not contained in $\mathrm{range}(A)$. Then $\mathrm{null}(CB) = \mathrm{null}(B)$, but now $\mathrm{range}(CB)$ is not contained in the range of $A$, since the range of $CB$ is exactly $W$. So you couldn't possibly be able to conclude that $\mathrm{range}(B)\subseteq\mathrm{range}(A)$ always holds in the first place.

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