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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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Think about straight lines through the origin in $\Bbb R^2$. – David Mitra May 8 '12 at 1:55
Straight lines are of dimension 1 in $R^2$ . It seems as my conclusion is wrong. – srijan May 8 '12 at 1:58
@srijan: You can now write up your observation as an answer (and accept it, if you'd like). This is explicitly encouraged by the SE network of sites; see here and here. – Zev Chonoles May 8 '12 at 2:00
Take spaces A and B of non col linear straight lines passing through origin in $R^2$ though $dim A = dim B$ ,but we cant conclude that A⊆B. They have only one point in common that is zero vector. Am i right sir? – srijan May 8 '12 at 2:12
@srijan: Simpler two say that A and B are two distinct lines that go through the origin in $\mathbb{R}^2$. Note that "collinear" is something we say about points, not about lines. – Arturo Magidin May 8 '12 at 2:21

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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