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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!

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    $\begingroup$ $V=U+W$ means every $v$ in $V$ can be written as a sum, $v=u+w$, with $u$ in $U$ and $w$ in $W$. $V=U\oplus W$ means $V=U+W$ and moreover the expression for each $v$ in $V$ is unique. $\endgroup$ – Gerry Myerson Sep 24 '16 at 6:34
  • $\begingroup$ I think that's what I was trying to say. But my main issue is applying it. Where (or better, how) does the P(1 - P) = 0 come into play? How does it (the P equation) figure into the direct sum? $\endgroup$ – Asinine Sep 24 '16 at 8:45
  • $\begingroup$ I'm not sure how $P^2=P$ comes into it (for one thing, you haven't told me what $P$ is – does it have some relation to $T$?). $\endgroup$ – Gerry Myerson Sep 24 '16 at 8:54
  • $\begingroup$ Yeah, okay so I'm stumped for the same reason. I found two solutions, both using $P$ -- and I don't remember either using any explanation aside from "try adding to it, multiplying by it, and mixing both those options together". I thought I'd bookmarked it, but apparently I haven't. Some of this stuff is way over my head so I could have easily missed some sort of association with $T$ or whatnot. $\endgroup$ – Asinine Sep 26 '16 at 1:14
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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.

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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

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  • $\begingroup$ I don't really see any connection between direct sums and xor, and that leads me to believe that you've misunderstood the definition. $\endgroup$ – Omnomnomnom Sep 30 '16 at 12:44
  • $\begingroup$ Say you have two vector spaces, given by $x^2$ and $x*(x+1)$. The venn diagram of each would overlap for those squares that are even numbers (e.g. 4, 16, 64, etc). So the direct sum of these two spaces would be all even numbers plus all odd squared numbers. Now that I can visualise how each space works, and their union, the explanation you provided below makes complete sense. I was lacking the tie between the concept and the understanding. $\endgroup$ – Asinine Sep 30 '16 at 12:54
  • $\begingroup$ To be perfectly frank, I can make no sense out of the first 3 sentences in that comment. The sets that you describe are definitely not vector spaces. You are correctly finding the symmetric difference of those two sets, but that does not connect to direct sums, so far as I can tell. $\endgroup$ – Omnomnomnom Sep 30 '16 at 13:02
  • $\begingroup$ I can see how + for vector spaces works a lot like the union of sets (that is, like "or"). Similarly, $\cap$ works a lot like the intersection of sets, or like "and". However, I wouldn't consider $\oplus$ an "operation" to be applied to two subspaces. Rather, I would say that $V=U \oplus W$ to be a special way to write the statement "$V= U+W$ and $U \cap W = \{0\}$". $\endgroup$ – Omnomnomnom Sep 30 '16 at 13:06
  • $\begingroup$ In your set theory metaphor, I would say that the integers are "the direct sum of the even numbers and the odd numbers" because the union of the odds and evens is the entire set, and no number is both odd and even. $\endgroup$ – Omnomnomnom Sep 30 '16 at 13:10

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