Prove a dimension relation $\dim S+\dim T<\dim X+\dim(\overrightarrow{S}+\overrightarrow{T})$ when $S\cap T=\emptyset$

Question

The proposition is:

Let $X$ be a finite-dimensional affine space and $S,T$ be subspaces of $X$. If $S\cap T=\emptyset$, then $$\dim S+\dim T<\dim X+\dim(\overrightarrow{S}+\overrightarrow{T}),$$ where $\overrightarrow{S}$ and $\overrightarrow{T}$ represents the direction of $S$ and $T$, respectively.

My attempt is as follows. Since $\dim S=\dim\overrightarrow{S}, \dim T=\dim\overrightarrow{T}, \dim X=\dim\vec X$, if we could show $\dim(\overrightarrow{S}+\overrightarrow{T})>\dim(\overrightarrow{S}\cap\overrightarrow{T})$, the inequality would be proved by $\dim(\overrightarrow{S}+\overrightarrow{T})\le\dim\vec X$ and the Grassmann relation $\dim(\overrightarrow{S}+\overrightarrow{T})=\dim\overrightarrow{S}+\dim\overrightarrow{T}-\dim(\overrightarrow{S}\cap\overrightarrow{T})$. However, I cannot establish $\dim(\overrightarrow{S}+\overrightarrow{T})>\dim(\overrightarrow{S}\cap\overrightarrow{T})$ from condition $S\cap T=\emptyset$. Worse still, I later realized that even if $S\cap T=\emptyset$, their directions $\overrightarrow{S}$ and $\overrightarrow{T}$ could be the same if $S\parallel T$. Therefore $\dim(\overrightarrow{S}+\overrightarrow{T})=\dim(\overrightarrow{S}\cap\overrightarrow{T})$ in that special case. Could you please help me with the proof (you don't have to follow my unsuccessful attempt)? Thank you very much.

Answer 1

My previous attempt is not in the right direction. Following is the proof I have just figured out.

The inequality is equivalent to $\dim S+\dim T-\dim(\overrightarrow{S}+\overrightarrow{T})<\dim X$ where the left side is equal to $\dim(\overrightarrow{S}\cap\overrightarrow{T})$ due to the Grassmann relation. If we can show $\dim(\overrightarrow{S}+\overrightarrow{T})<\dim X$, then $\dim(\overrightarrow{S}\cap\overrightarrow{T})\le\dim(\overrightarrow{S}+\overrightarrow{T})<\dim X$, which establishes the proposition.

To this end, we need to make use of two facts:

1. Let $a\in S$ and $b\in T$. Then $S\cap T\ne\emptyset$ is equivalent to $\overrightarrow{ab}\in\overrightarrow{S}+\overrightarrow{T}$.
2. Let $a\in S$ and $b\in T$. $\overrightarrow{\langle S\cup T\rangle}=\overrightarrow{S}+\overrightarrow{T}+K\vec{ab}$ where $K$ is the field of scalars.

Fact 1. is not hard to prove. A proof of Fact 2. is here.

From fact 1., we can infer that $\overrightarrow{S}+\overrightarrow{T}$ is independent of $K\vec{ab}$. To show this, if $(\overrightarrow{S}+\overrightarrow{T})\cap K\vec{ab}$ contains an element $c\vec{ab}\ne 0$, then $c\ne0$, so we have $\vec{ab}\in (\overrightarrow{S}+\overrightarrow{T})$, which is forbidden by Fact 1., because $S\cap T=\emptyset$. Taking dimension on both sides in Fact 2., we have $\dim\overrightarrow{\langle S\cup T\rangle}=\dim\bigl((\overrightarrow{S}+\overrightarrow{T})+K\vec{ab}\bigr)$. The vector spaces in the parentheses on the right side are independent as we have just shown, so we have $\dim\overrightarrow{\langle S\cup T\rangle}=\dim(\overrightarrow{S}+\overrightarrow{T})+\dim K\vec{ab}=\dim(\overrightarrow{S}+\overrightarrow{T})+1$. $\dim\overrightarrow{\langle S\cup T\rangle}\le\dim\vec X$ in general, so we finally get $\dim(\overrightarrow{S}+\overrightarrow{T})<\dim\vec X$.