Finite Dimensional Vector Space and its subspaces

Question

Here is the problem.

Suppose $V$ is a finite dimensional vector space and $U, W$ are subspaces of $V$. ($U, W \leq V$) Also let $\mathfrak{B}, \mathfrak{C}$ be a basis of $U, W$, respectively.

$\quad $Prove that if $U\cap W = 0$ (zero vector space), $\mathfrak{B} \cap \mathfrak{C} = \emptyset$ and $\mathfrak{B} \cup \mathfrak{C}$ is a basis of $U+W$.

So I tried by letting $\mathfrak{B}, \mathfrak{C}$ be a basis of given vector spaces $U$ and $W$. It really occurs to me that I should be using something related to $U\cap W = 0$ (zero vector space) but I can't really get anything out of this. So I cannot get a grasp of where to start from.

It feels like I might have to use the Basis Extension Theorem. But I also don't really get why this might be true. Can I get a hint on this or at least a intuitive reason for why this might be true?

Thanks in advance.

Answer 1

First, note that $0\notin\mathfrak{B}$ and $0\notin\mathfrak{C}$, and $$\mathfrak{B}\cap\mathfrak{C}\subset U\cap W=\{0\}.$$ It follows that $\mathfrak{B}\cap\mathfrak{C}=\emptyset$.

To show that $\mathfrak{B}\cup\mathfrak{C}$ is a basis, you need to show that

i) $\mathfrak{B}\cup\mathfrak{C}$ spans $U+W$, and

ii) $\mathfrak{B}\cup\mathfrak{C}$ is linearly independent.

To show (i), note that any $v\in U+W$ can be written as $v=u+w$ for some $u\in U$ and $w\in W$. Well, $$u=\sum_{b\in\mathfrak{B}}\lambda_bb\;\;\;\mbox{and}\;\;\;w=\sum_{c\in\mathfrak{C}}\mu_cc$$ for some $\lambda_b,\mu_c\in F$. Therefore, $$v=u+w=\sum_{b\in\mathfrak{B}}\lambda_bb+\sum_{c\in\mathfrak{C}}\mu_cc\in\mathrm{Span}(\mathfrak{B}\cup\mathfrak{C})$$ showing that $\mathfrak{B}\cup\mathfrak{C}$ spans $U+V$.

To show (ii), assume $$ \sum_{b\in\mathfrak{B}}\lambda_bb+\sum_{c\in\mathfrak{C}}\mu_cc=0. $$ Then, $$\sum_{b\in\mathfrak{B}}\lambda_bb=-\sum_{c\in\mathfrak{C}}\mu_cc. $$ The left hand side of the equation above belongs to $U$, while the right hand side belongs to $W$. Hence, both sides belong to $U\cap W=\{0\}$. That is, $$\sum_{b\in\mathfrak{B}}\lambda_bb=\sum_{c\in\mathfrak{C}}\mu_cc=0. $$ Since $\mathfrak{B}$ is linearly independent, all $\lambda_b=0$ and, since $\mathfrak{C}$ is linearly independent, all $\mu_c=0$. Hence $\mathfrak{B}\cup\mathfrak{C}$ is linearly independent, as required.

Answer 2

A sketch of a proof . $\mathfrak{B}$ and $\mathfrak{C}$ are subsets of $U$ and $W,$ respectively, so if $U$ and $W$ are disjoint, so are $\mathfrak{B}$ and $\mathfrak{C}.$ Again since $U$ and $W$ are disjoint, any vector of $U+W$ has a unique expression as $u+w,$ where $u\in U$ and $w \in W:$ if $u+w=u'+w',$ then $u-u'=w-w',$ which belongs to both $U$ and $W,$ so must be the zero vector, whence $u=u'$ and $w=w'.$ Consequently, $u+w=0$ implies $u=w=0,$ $0$ being the zero vector.

For an arbitrary $u+w,$ $u,$ (resp. $w$) is a unique linear combination of the vectors in $\mathfrak{B},$ (resp. $\mathfrak{C}),$ so $\mathfrak{B} \cup \mathfrak{C}$ spans $U+W.$ That this set is linearly independent can be shown using the argument at the end of the first paragraph.