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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.

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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 $W_1+W_2=W_2$.

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

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  • $\begingroup$ Sir, please elaborate last line "so there aren't many subspaces. $\endgroup$ – Akash Patalwanshi Mar 22 '19 at 13:29
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    $\begingroup$ $F$ 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 $F_2$? $\endgroup$ – Dirk Mar 22 '19 at 14:25
  • $\begingroup$ Sir, lf $F$ is one dimensional vector space then $F$ and $\{0\}$ are only subspaces of $F$. $\endgroup$ – Akash Patalwanshi Mar 22 '19 at 16:41
  • $\begingroup$ 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. $\endgroup$ – Akash Patalwanshi Mar 22 '19 at 16:55
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$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$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 $W_{1} = W_{1} \cap W_{2}$.

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

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