Dimension of sum and intersection of vector space.

I am trying to understand the proof of the following: Suppose $U,W$ are vector subspace of $V$, then $\dim (U+W)+\dim (U \cap W)= \dim (U) +\dim (W).$

The proof goes like this: Let $S: V \rightarrow V/W$ be the natural surjection. Then we have $\dim (V) = \dim (W) +\dim (V/W)$ by rank-nullity. Now let $T: U \rightarrow (U+W)/W$. Then we have $\dim \ker (T)= \dim (U \cap W)$ and $\dim Im (T)= \dim (U+W) - \dim (W)$ and the result follows.

Basically I don't understand what is going on after "Now let $T$..." I don't understand how is $T$ actually define, and why the rank and nullity of $T$ equals that (which I think will be clear once I know how $T$ is defined), could someone please help, thanks!

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3 Answers

$T$ is defined as the composition $U\to U+W\to (U+W)/W$, where the first map is just the inclusion, and the second map is the canonical projection.

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Looks like $T$ is the inclusion of $U$ into $U + W$ followed by the quotient map to $(U + W)/W$.

In this interpretation, an element $u$ of $U$ maps to $0$ if and only if $u \in W$, so $\ker(T) = U \cap W$. Further, $T$ is (more-or-less obviously) surjective, so $\dim \operatorname{Im}(T) = \dim(U + W)/W = \dim(U + W) - \dim W$.

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$U+W$ is the subspace of $V$ that consists of vectors of the form $u+w$, where $u\in U$, $w\in W$. As $W$ is a subspace of $U+W$, one can consider the quotient $(U+W)/W$, which is the space of all vectors $(u+w) + W$.

Now, $T$ is the linear function that takes $u$ to the class of $u$ in this quotient, which is $u+W$. If $u$ is in the kernel of $T$, it means that $u+W = W$, i.e that $u\in W$, hence $\ker(T) \subseteq U\cap W$, and the other inclusion is true also, since if $u\in U\cap W$, $u\in W$ and $u+W = W$. We have proved $\ker(T) = U\cap W$.

As for $\mathrm{im}(T)$: $T$ is surjective, because if $(u+w)+W\in (U+W)/W$, this vector equals $u+W$, and $u$ is a preimage for this. So $\dim \mathrm{Im}(T) = \dim (U+W)/W = \dim (U+W) - \dim W$ as you already now (it works as your first equality, with $S$).

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