# Direct Sum of Subspaces - Question

• Jun 14th 2011, 05:59 AM
worc3247
Direct Sum of Subspaces - Question
Let U and W be subspaces of V . Show that the following are equivalent:
(i) for each vector $\displaystyle v \in V$ there exist unique vectors $\displaystyle u \in U$ and $\displaystyle w \in W$ such that v=u+w;
(ii) dim V $\displaystyle \leq$ dim U + dim W and $\displaystyle U\cap W =$ {0}.

I am currently stuck showing i) implies ii). So far all I have is that Sp(U+W)=V, but can't see how this will help. Any tips on where to start?
• Jun 14th 2011, 06:30 AM
Deveno
Re: Direct Sum of Subspaces - Question
suppose B = {u1,u2,...,um} is a basis for U, and C = {w1,w2,...,wn} is a basis for W.

is BUC a spanning set for U+W? is the size of a basis for V:

a) equal to, b) less than, c) greater than, d) less than or equal to, or e) greater than or equal to a spanning set for V?

once you have answered THAT question, what can we say about U∩W if B∩C = Ø?

why does the uniqueness of v = u+w imply something about the intersection of the two bases?

(hint: write the 0-vector in this form).
• Jun 14th 2011, 07:07 AM
worc3247
Re: Direct Sum of Subspaces - Question
Ok, by writing u and v as linear combinations of the elements of B and C, I can show that Sp(BUC)=U+W, and I know that Sp(U+W)=V. So from this (I think) I can then say Sp(BUC)=V. Now intuitively I can see that the inequality is correct, but how do I write it formally?
• Jun 14th 2011, 05:19 PM
Deveno
Re: Direct Sum of Subspaces - Question
again, when you have a basis A for a vector space V, dim(V) = |A|. that is how you know the dimension, it's the size of any basis.

now if a set is a spanning set, it contains a basis (but it may be too big, because there may be some linearly dependency).

if a set is linearly independent you can extend it to a basis (because it may be too small, and does not span).

the concepts are stacked like this: linearly independent set ≤ basis ≤ spanning set.

so, since BUC is a spanning set for V, we have dim(V) ≤ |BUC| = |B| + |C| - |B∩C| ≤ |B| + |C| = dim(U) + dim(W)

(the middle equality for |BUC| is also known as the inclusion-exclusion principle: we count elements common to both sets twice, so we have to subtract them out).

saying sp(U+W) = V is overkill. we already know U+W = V (since we can write every v as u+w). certainly every vector space spans itself: sp(V) = V.

but generally, you want to do MUCH better than that, you want to find as small a spanning set as possible. such a set is called a basis. BUC is a LOT smaller than U+W.

and smaller spanning sets are better (just like bigger linearly independent sets are better). if BUC spans U+W, then it certainly spans V

since V EQUALS U+W. and since B is a basis for U, and C is a basis for W, |B| = dim(U), |C| = dim(W). isn't that what we're after? trying to express

some sense of relative size between the dimensions? when ever you see a question involving "dimension" you should think: "i need a basis".

now, we know that the 0-vector is in V. and since 0 = 0+0, and 0 is in U, and 0 is in W, this must be the ONLY way we can write 0 as a sum of something

in U, and something in W. so using our bases B and C, suppose we have 0 = a1u1 + a2u2 +...+ amum + b1w1 + b2w2 +....+ bnwn.

then a1u1 + a2u2 +...+ amum and b1w1 + b2w2 + ...+ bnwn have to be both 0 (why? the answer is above).

but that means that BUC is a _____ ______ set, which means that dim(U+W) = _______ + ________ ,

which means in particular that |B∩C| = _______ , which in turn means that U∩W = ______ (because it has an _____ basis).
• Jun 20th 2011, 05:04 PM
obd2
Re: Direct Sum of Subspaces - Question
Ok. first (i) $\displaystyle \Rightarrow$Given (i) then you know that $\displaystyle V \subseteq U \oplus W$. We also know that $\displaystyle dim(U\oplus W) = dim(U) + dim(W)$. It is known that the dimesion of a subspace is lesser than or equal to the dimesion of the larger space. Hence we have shown that
$\displaystyle $$dim(V) \le dim(U\oplus W) = dim(U ) + dim(W)$$$
• Jun 20th 2011, 05:19 PM
obd2
Re: Direct Sum of Subspaces - Question
Please excuse my previous post as I accidentally pressed the submit button instead of review button. English is not my first language but I'm doing my best. I hope that the following helps, I like to avoid using any vectors whatsoever in proofs like this.

Ok. first (i) $\displaystyle \Rightarrow$(ii). Given (i) then you know that $\displaystyle V \subseteq U \oplus W$. We also know that $\displaystyle dim(U\oplus W) = dim(U) + dim(W)$. It is known that the dimesion of a subspace is lesser than or equal to the dimesion of the larger space. Hence we have shown that
$\displaystyle $$dim(V) \le dim(U\oplus W) = dim(U ) + dim(W)$$$.
Now $\displaystyle (ii) \Rightarrow (i)$. Given (ii) we automatically know that
$\displaystyle $$dim(V) \le dim(U) + dim(W)$$$, and we know that $\displaystyle U\cap W = \{0\}$. Next, we know that since $\displaystyle U$ and $\displaystyle W$ are both subspaces of $\displaystyle V$ then their sum (that is $\displaystyle U+W$) is also a subspace of $\displaystyle V$, hence the following holds:
$\displaystyle $$dim(U+W) \le dim(V). Finally, since we have the equation \displaystyle dim(U+W) = dim(U) + dim(W) - dim(U\cap W), which gives us that \displaystyle dim(U+W) = dim(U) + dim(W) since \displaystyle dim(U\cap W) = dim(\{0\}) = 0. Effecetively we have shown that \displaystyle$$dim(V) \ge dim(U) + dim(W)$$also holds. Then we must have \displaystyle dim(V) = dim(U) + dim(W) = dim(U + W) . This gives us that \displaystyle V = U+W and, since \displaystyle U\cap W = \{0\}, therefore \displaystyle V = U \oplus W which is equivalent to (i). • Jun 20th 2011, 05:23 PM Also sprach Zarathustra Re: Direct Sum of Subspaces - Question Quote: Originally Posted by obd2 Please excuse my previous post as I accidentally pressed the submit button instead of review button. English is not my first language but I'm doing my best. I hope that the following helps, I like to avoid using any vectors whatsoever in proofs like this. Ok. first (i) \displaystyle \Rightarrow(ii). Given (i) then you know that \displaystyle V \subseteq U \oplus W. We also know that \displaystyle dim(U\oplus W) = dim(U) + dim(W). It is known that the dimesion of a subspace is lesser than or equal to the dimesion of the larger space. Hence we have shown that \displaystyle$$dim(V) \le dim(U\oplus W) = dim(U ) + dim(W)$$. Now \displaystyle (ii) \Rightarrow (i). Given (ii) we automatically know that \displaystyle$$ dim(V) \le dim(U) + dim(W)$$, and we know that \displaystyle U\cap W = \{0\}. Next, we know that since \displaystyle U and \displaystyle W are both subspaces of \displaystyle V then their sum (that is \displaystyle U+W) is also a subspace of \displaystyle V, hence the following holds: \displaystyle$$ dim(U+W) \le dim(V)$.
Finally, since we have the equation $\displaystyle dim(U+W) = dim(U) + dim(W) - dim(U\cap W)$, which gives us that $\displaystyle dim(U+W) = dim(U) + dim(W)$ since $\displaystyle dim(U\cap W) = dim(\{0\}) = 0$. Effecetively we have shown that
$\displaystyle $$dim(V) \ge dim(U) + dim(W)$$$
also holds. Then we must have $\displaystyle dim(V) = dim(U) + dim(W) = dim(U + W)$. This gives us that $\displaystyle V = U+W$ and, since $\displaystyle U\cap W = \{0\}$, therefore $\displaystyle V = U \oplus W$ which is equivalent to (i).

Next time you can just press on 'Edit post'...