We have two subspaces $A$ and $B$ of a vector space $V$ such that $\dim A\leq \dim B$. Can we conclude that $A\subseteq B $ ? I need a proper justification.
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No, you cannot conclude that. It is true that if $A\subseteq B$, then $\dim A\leq \dim B$, but the converse does not hold. For example, in $\mathbb{R}^3$, the $z$-axis, $A=\{(0,0,z)\mid z\in\mathbb{R}\}$ is a 1-dimensional subspace; the $xy$-plane, $B=\{(x,y,0)\mid x,y\in\mathbb{R}\}$ is a 2-dimensional subspace; but $A$ is not contained in $B$: in fact, their intersection is just the origin. You can put a lower bound in the intersection using the fact that $$\dim(A+B) = \dim(A)+\dim(B) - \dim(A\cap B)$$ and that $\dim(A+B)\leq\dim(V)$. From this, we can conclude that $$\dim(A)+\dim(B)-\dim(A\cap B)\leq\dim(V),$$ and therefore that $$\dim(A\cap B) \geq \max\Bigl( 0, \dim(A)+\dim(B)-\dim(V)\Bigr),$$ but you can't say more than that. In particular, the only times that you can guarantee that $A\subseteq B$ is when $\dim(B)=\dim(V)$, because in that case you have $B=V$ and so $A\subseteq B$ holds; and when $\dim(A)=0$, because in that case we have $A=\{0\}$ so $A\subseteq B$ will hold. |
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