Let $V=W_1\oplus W_2=U_1\oplus U_2$, if $W_1\subset U_1$ and $ W_2\subset U_2$ $\Rightarrow$ $W_1=U_1$ and $W_2=U_2$

Question

I'm trying to show this:

Let be V a finite dimensional space and $V=W_1\oplus W_2=U_1\oplus U_2$, if $W_1\subset U_1$ and $ W_2\subset U_2$ $\Rightarrow$ $W_1=U_1$ and $W_2=U_2$

I was given a hint that I should view it by showing that:

$W_1\subset U_1$ and$ W_2\subset U_2$ $\Rightarrow$ $Dim(W_1)\le Dim (U_1)$, $Dim(W_2)\le Dim (U_2)$ (Where $Dim$ stands for dimension) and showing that $Dim(W_1)= Dim (U_1)$ , $Dim(W_2)= Dim (U_2)$

because if we consider $Dim(W_1) \lt Dim (U_1)$ and $Dim(W_2) \lt Dim (U_2)$ we'll get a contradiction with $V=W_1\oplus W_2=U_1\oplus U_2$

What I think the contradiction will be is in: $Dim(W_1\oplus W_2) \lt Dim(U_1\oplus U_2)$. But is this enough to prove what I want?

Thanks!

Answer 1

Let $u_1\in U_1\subset V=W_1\oplus W_2$, we may write $u_1=w_1+w_2$ for some $w_1\in W_1$ and $w_2\in W_2.$ Then $$u_1-w_1=w_2\in W_2\subset U_2.$$ Also, since $W_1\subset U_1$, we have $u_1-w_1\in U_1$ or $w_2\in U_1$. Thus we have $w_2={\it 0}$ because of the assumption that $U_1\cap U_2=\{{\it 0}\,\}$. Hence $u_1=w_1\in W_1$ and therefore we conclude that $U_1\subset W_1$, which implies $U_1=W_1$. The same argument for $U_2=W_2$.