(a) Let $W_{1}$ and $W_{2}$ be subspaces of a vector space $V$ such that $V = W_{1}\oplus W_{2}$. If $\mathcal{B}_{1}$ and $\mathcal{B}_{2}$ are basis for $W_{1}$ and $W_{2}$, respectively, show that $\mathcal{B}_{1}\cap\mathcal{B}_{2} = \varnothing$ and $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ is a basis for $V$.
(b) Conversely, let $\mathcal{B}_{1}$ and $\mathcal{B}_{2}$ be disjoint basis for $W_{1}$ and $W_{2}$, respectively, of a vector space $V$. Prove that if $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ is a basis for $V$, then $V = W_{1}\oplus W_{2}$.
MY ATTEMPT
(a) Since $V = W_{1}\oplus W_{2}$, given $v\in V$, it can be (uniquely) written as $v = w_{1} + w_{2}$, where $w_{1}\in W_{1}$ and $w_{2}\in W_{2}$.
If we denote a basis of $W_{1}$ by $\mathcal{B}_{1} = \{w_{11},w_{12},\ldots,w_{1m}\}$ and a basis of $W_{2}$ by $\mathcal{B}_{2} = \{w_{21},w_{22},\ldots,w_{2n}\}$, we conclude that $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ spans $V$. Since $\dim V = \dim W_{1} + \dim W_{2}$, it results that $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ is a basis for $V$.
In order to support such claims, we have made use of the following proposition:
Given a finite-dimensional vector space $V$ as well as two subspaces $W_{1}$ and $W_{2}$ of it, one has that \begin{align*} \dim(W_{1}) + \dim(W_{2}) = \dim(W_{1} + W_{2}) + \dim(W_{1}\cap W_{2}) \end{align*}
as well as
If the set of vectors $\mathcal{B}\subset V$ spans $V$ and $|\mathcal{B}| = \dim V$, then $\mathcal{B}$ is a basis for $V$, where $V$ a finite dimensional vector space.
and we are done with the second part of the problem. Let us now prove that $\mathcal{B}_{1}\cap\mathcal{B}_{2} = \varnothing$.
Indeed, if it was not the case, we should have a vector $v\in\mathcal{B}_{1}\cap\mathcal{B}_{2}$ which is not the null vector.
Since $\text{span}\{v\}\subseteq\text{span}(\mathcal{B}_{1}\cap\mathcal{B}_{2})\subseteq\text{span}(\mathcal{B}_{1}) = W_{1}$ and $\text{span}\{v\}\subseteq\text{span}(\mathcal{B}_{1}\cap\mathcal{B}_{2})\subseteq\text{span}(\mathcal{B}_{2}) = W_{2}$, we conclude that $v\in W_{1}\cap W_{2}$, which is impossible because $W_{1}\cap W_{2} = \{0\}$. Thus $\mathcal{B}_{1}\cap\mathcal{B}_{2} = \varnothing$.
(b) Once again, we shall assume that $\dim V = k$, $\dim W_{1} = m$ and $\dim W_{2} = n$.
Since $\mathcal{B}_{1}\cap\mathcal{B}_{2} = \varnothing$ and $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ is a basis for $V$, any vector $v\in V$ can be written as \begin{align*} v = a_{1}w_{11} + a_{2}w_{12} + \ldots + a_{m}w_{1m} + b_{1}w_{21} + b_{2}w_{22} + \ldots + b_{n}w_{2n} = w_{1} + w_{2} \end{align*} where $w_{1}\in W_{1}$ and $w_{2}\in W_{2}$. Consequently, $V = W_{1} + W_{2}$.
Besides, $W_{1}\cap W_{2} = \{0\}$. Otherwise, we would have a contradiction.
Precisely speaking, let us assume there exists $v\neq 0$ and $v\in W_{1}\cap W_{2}$. Thus $v\in W_{1}$ and $v\in W_{2}$. Therefore we should have \begin{align*} v & = c_{1}w_{11} + c_{2}w_{12} + \ldots + c_{m}a_{1m} = d_{1}w_{21} + d_{2}w_{22} + \ldots + d_{n}w_{2n}\\\\ & \Rightarrow c_{1}w_{11} + c_{2}w_{12} + \ldots + c_{m}a_{1m} - d_{1}w_{21} - d_{2}w_{22} - \ldots - d_{n}w_{2n} = 0\\\\ & \Rightarrow c_{1} = c_{2} = \ldots = c_{m} = d_{1} = d_{2} = \ldots = d_{n} = 0\\\\ & \Rightarrow v = 0 \end{align*} which is a contradiction. This is because the elements of $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ are LI, and we are done.