Show that there is a case that $\dim U + \dim W + \dim T = \dim V$, but $U + W + T \neq V$. Having the following problem. I need to show that there are exist some subspaces $U,W,T$ and finite-dimensional space $V$ that this case is true: $\dim U + \dim W + \dim T = \dim V$, but $U + W + T\neq V$.
My idea is following, I know how to prove that $\dim(+)+\dim(∩)=\dim +\dim $, then $∩=\{0\}$ and thinking of using it for proving the statement from above but not sure if it can be used in case of three subspaces. Can you please let me know if this is the right way or there is another idea for proving this.
Thank you.
 A: Let us analyze the “more interesting” case (as suggested by David Mitra): $U,W,T\subset\pi\subset V$, where $V$ is a vector space of dimension $3$, $\pi$ is a plane (space of dimension $2$) and $U,W,T$ are three (distinct) lines (spaces of dimension $1$) all contained in $\pi$, i.e., coplanar.
It is clear that $U+W+T\neq V$, since $U+W+T$ is the smallest subspace of $V$ that contains all $U, V$ and $T$, and $\pi$ (which is strictly smaller than $V$) does contain all three. However, $$\dim U+\dim W+\dim T=1+1+1=3=\dim V.$$
In order to see this in the light of the formula you mention, note that $U+W+T=(U+W)+T$, hence
$$\dim((U+W)+T)=\dim(U+W)+\dim T-\dim((U+W)\cap T),$$
and also that
$$\dim(U+W)=\dim U+\dim W-\dim(U\cap W)$$
Substituting the latter equation into the former, we get:
$$\dim((U+W)+T)=\dim U+\dim W+\dim T-\dim((U+W)\cap T)-\dim(U\cap W)$$
Let us assume the lines are distinct. In this case, $U\cap W=\{0\}$, hence $\dim(U\cap W)=0$. Lastly, $U+W=\pi$ (since they are distinct thus generated by two independent vectors), hence $(U+W)\cap T=T$, and $\dim((U+W)\cap T)=1$. Plugging this in our equation, we get:
$$\dim((U+W)+T) =1+1+1-1-0=2,$$
therefore $U+W+T$ cannot be equal to $V$.
