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?
If $null(A) \subset null(B)$ can we draw any conclusion about range spaces of A and B
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0Doesn't matter. You still can't get the result. – 2012-05-08
1 Answers
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.