The direct sum of a rank and null space

Question

Firstly, I have no idea how to use most of the fancy mathjax formatting stuffs yet. I'm currently too frustrated trying to figure out how to understand this problem (due in 10 hours), so please feel free to refer anything for that while answering. :)

I've got a hw problem stated as such: Let $V$ be a finite-dimensional vector space and T:V->V be linear.
A) Suppose that V = R(T) + N(T). Prove that V = R(T) $\oplus$ N(T).
B) Suppose that R(T) $\cap$ N(T) = {$0$}. Prove that V = R(T) $\oplus$ N(T).

I've found two separate answer explanations for part A (both along the lines of $P$ = $P^2$ with $P$(1 - $P$) = 0), but I don't understand why that is the answer. The largest problem is probably my confusion with how direct sums work. I understand that the subspace $V$ $\oplus$ subspace $W$ = a vector space, when
- 1) $V$ $\nsubseteq$ $W$ (I think that's the correct notation? Basically all members of V and all of W will not be equal/overlap), and when
- 2) $V$ $\cap$ $W$ = {$0$} (i.e. they only have the zero vector in common)

But I'm having issues correlating those two points to anything outside of the definition, so understanding and proving this is completely foreign to me. I'd love answers even after this is due, since I suspect this class will be heavy on understanding these concepts.

Thanks in advance!

Answer 1

The easiest explanation I got for a direct sum was the Venn diagram, and the "generic" programming idea of xor. Basically, it's the two vector spaces summed together, with their intersection subtracted.

Because the only "common denominator" of sorts (only common ground, I suppose?) for $N(T)$ and $R(T)$ is the null vector, the proof lies in explaining that $N(T)$ is indeed $\{0\}$.

I found this incredibly useful to visualise the issue: https://en.wikipedia.org/wiki/Exclusive_or

Answer 2

Here's a useful fact: for any spaces $U$ and $W$, we have $$ \dim (U + W)= \dim U + \dim W - \dim(U\cap W) $$ In this case, take $U$ to be the range and $W$ to be the nullspace. By the rank-nullity theorem, we know that $$ \dim(U) + \dim(W)= \dim (V) $$

Following Gerry's definition, the two requirements for the direct sum are

• $ V = U + W$
• $U \cap W = \{0\}$

Or, to put it another way, to check that $V= U \oplus W$, it suffices to check that both of the following are true:

• $\dim(U + W)= \dim(V)$
• $\dim(U\cap W)= 0$

Now, we can consider the question. Suppose that $\dim(U+W) = \dim(V)$. Use the first and second equations to show that we must also have $\dim(U\cap W)=0$, which would mean that $V=U \oplus W$ as desired.

The answer to the second part is similar.