Show that $W_1+W_2$ is one of the subspaces and $W_1\cap W_2 $ is equal to other 
Question:  given $W_1$, $W_2$ be finite dimensional subspaces of vector space such that $$\dim(W_1+W_2)=1+\dim(W_1\cap W_2)$$
Show that $W_1+W_2$ is equal to one of the subspaces and $W_1\cap W_2$ is equal to the other.

I know that,
\begin{align}\dim(W_1\cap W_2)&≤\dim W_1\\
\dim(W_1\cap W_2)&≤\dim W_2 \\
\dim(W_1+W_2)&=\dim W_1+ \dim W_2-\dim(W_1\cap W_2)\end{align}
With this information,
$$1+\dim(W_1\cap W_2)=\dim W_1 +\dim W_2-\dim(W_1\cap W_2)$$
Hence $$2\dim(W_1\cap W_2)= \dim W_1+\dim W_2-1$$
Still unable to prove, please help.
 A: The spaces $W_1$ and $W_2$ both contain $W_1 \cap W_2$, and are both contained in $W_1 + W_2$. Now the factor spaces $F_1 := W_1/(W_1 \cap W_2)$ and $F_2 := W_2/(W_1\cap W_2)$ are both subspaces of $(W_1+W_2)/(W_1\cap W_2) := F$. But this space $F$ has dimension one by what you are given, so there aren't that many subspaces.
A: $\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$Hint 1

Suppose $W_{1} + W_{2} \ne W_{1}$. Then there is $u \in W_{2} \setminus W_{1}$.

Hint 2

Then $W_{1} + W_{2} = (W_{1} \cap W_{2}) + \Span{u}$, by the dimension condition.

Hint 3

Then $W_{1} + W_{2} = (W_{1} \cap W_{2}) + \Span{u} \subseteq W_{2}$, so that $W_{1} + W_{2} = W_{2}$.

Hint 4

$W_{1} + W_{2} \supsetneq W_{1} \supseteq W_{1} \cap W_{2}$, so that by the dimension condition $W_{1} = W_{1} \cap W_{2}$.

A: So far so good with your solution. Assume now that both $W_1$ and $W_2$ are bigger than $W_1\cap W_2$, then you have that $\dim(W_1\cap W_2)+1\le \dim (W_1)$ and the same for $W_2$ and your last equation becomes 
\begin{align}2\dim(W_1\cap W_2)&=\dim W_1+\dim W_2-1\\&\ge (1+\dim(W_1\cap W_2))+(1+\dim(W_1\cap W_2))-1\\&=1+2\dim(W_1\cap W_2)\end{align} which is a contradiction. So, either $W_1=W_1\cap W_2$ or $W_2=W_1\cap W_2$. Say without loss of generality that this is $W_1$. Then $W_1\subseteq W_2$ and it follows that $W_1+W_2=W_2$.
