Let $V_1,V_2$ be subspaces of $V$. If $\dim(V_1+V_2)=\dim(V_1 \cap V_2) + 1$ then prove that $ V1 \subseteq V2 $ or $V2 \subseteq V1$.

Question

We know that $\dim(V_1+V_2)=\dim(V_1)+\dim(V_2)-\dim(V_1 \cap V_2)$, But $\dim(V_1+V2)=\dim(V1 \cap V2)+ 1$ (given),

Now if we assume that $V_1 \subseteq V_2$ , $V_1 \cap V_2= V_1$, Then at one side it will be $\dim(V_1)$ while other side it will be $\dim(V_1)+1$. How it is possible?

Answer 1

As for why this is possible, here is an example: Take $V_1=\mathbb{R}^2$ and $V_2=\{0\}\times \mathbb{R}$. Then you can check easily that all the statements are satisfied with $\dim V_2=1$.

For a hint for the prove: Try to extend a basis for $V_1\cap V_2$ to a basis of $V_1+V_2$. Imagine what would happen if none of the inclusions is satisfied.