# 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 $$2\dim(W_1\cap W_2)= \dim W_1+\dim W_2-1$$

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 $$W_1+W_2=W_2$$.

• Sir, $dim(W_1\cap W_2)-1≤dimW_1$ but sir, while using this in equation which I had obtained, you written, $dimW_1≥1+dim(W_1\cap W_2)$ – Akash Patalwanshi Mar 22 '19 at 13:33
• Ok, there was a typo. Should be correct now. Thanks for noticing – Jimmy R. Mar 22 '19 at 13:38

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.

• Sir, please elaborate last line "so there aren't many subspaces. – Akash Patalwanshi Mar 22 '19 at 13:29
• $F$ is a space of dimension one. Can you write down all subspaces of a space of dimension one? Which one of them can be $F_1$, which one can be $F_2$? – Dirk Mar 22 '19 at 14:25
• Sir, lf $F$ is one dimensional vector space then $F$ and $\{0\}$ are only subspaces of $F$. – Akash Patalwanshi Mar 22 '19 at 16:41
• Thank you so ...much sir, I think you want to say, either $F_1=F$ and $F_2=\{0\}$ or $F_2=F$ and $F_1=\{0\}$. In the first case, $W_1=W_1+W_2$ and $W_1\cap W_2=W_2$ and in second case $W_1+W_2=W_2$ and $W_1\cap W_2=W_1$. Beautiful answer. – Akash Patalwanshi Mar 22 '19 at 16:55

$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$$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 $$W_{1} = W_{1} \cap W_{2}$$.

• Thank you so much. Sir – Akash Patalwanshi Mar 22 '19 at 13:48