proving that two subspaces are direct sum

**Stormey** · January 6th 2013, 02:19 PM

Let $W,U\subset V$ be subspaces.

i need to prove that if $dim(V)=dim(U)+dim(W)$ and $W\cap U=\left \{ \vec0 \right \}$ then $W\oplus U=V$ .

here's the thing.
i proved it, but without considering $W\cap U=\left \{ \vec0 \right \}$ .
(i proved that if $dim(V)=dim(U)+dim(W)$ then $W\oplus U=V$ )

and that's obviously can't be, i mean there must be something i'm doing wrong, i just can't figure out what.

here's what i did:

to prove that U and W are the direct sum of V, i just need to show that for every $v\in V$ : there's only one linear combination of $u\in U$ and $w\in W$ .

so:
Let $\left \{ u_1, u_2... u_n \right \}$ be the basis for U.
Let $\left \{ w_1, w_2... w_m \right \}$ be the basis for W.
now, since $dim(V)=dim(U)+dim(W)$ , i can conclude that the number of vectors in V's basis must be $n+m$ .
so, the basis for V can now be:
$\left \{ u_1, u_2... u_n,w_1, w_2... w_m \right \}$

now, let's presume v can be presented in two different ways, and show that it's actually the same presentation, so:
let's presume there are scalars $a_1,a_2... a_{n+m}\in \mathbb{R}$ , not all 0, and $b_1,b_2... b_{n+m}\in \mathbb{R}$ , not all 0, such that:

$a_1u_1+a_2u_2+...+a_nu_n+a_{n+1}w_1+a_{n+2}w_2+... +a_mw_m=v$
$b_1u_1+b_2u_2+...+b_nu_n+b_{n+1}w_1+b_{n+2}w_2+... +b_mw_m=v$

if we'll subtruct one from another we'll get:

$(a_1-b_1)u_1+(a_2-b_2)u_2+...+(a_{n}-b_{n})u_n+(a_{n+1}-b_{n+1})w_1+(a_{n+2}-b_{n+2})w_2+...+(a_m-b_m)w_m=0$

and since $\left \{ u_1, u_2... u_n,w_1, w_2... w_m \right \}$ are linear independent (basis for V), then $a_1=b_1$ , $a_2=b_2$ and so on...
so that's all.

my question now is: where exactly does $W\cap U=\left \{ \vec0 \right \}$ fit in? why do i even need it here?

it's quite a task to use the math terminology when it's not your native language, so i hope i used it right and everything is clear enough...

thanks in advanced!

**GJA** · January 6th 2013, 04:21 PM

Hi Stormey,

The argument uses $U\cap W=\{0\}$ when you conclude that $\{u_{1},\ldots, u_{n},w_{1},\ldots,w_{m}\}$ is a basis for $V.$ Just because $\{u_{1},\ldots, u_{n}\}$ and $\{w_{1},\ldots, w_{m}\}$ are linearly independent sets of vectors on their own does not always mean $\{u_{1},\ldots, u_{n},w_{1},\ldots, w_{m}\}$ must be a linearly independent collection too. For example, the sets $\{[1,0,0], [0,1,0]\}$ and $\{[0,1,0], [0,0,1]\}$ are each linearly indepdent sets of vectors on their own, but $\{[1,0,0], [0,1,0], [0,1,0], [0,0,1]\}$ is not a linearly indepedent set of vectors.

Does this answer your question? Let me know if anything is unclear. Good luck!

**Stormey** · January 6th 2013, 10:09 PM

Hi GJA, and thanks for your help.

i'm aware that if two subspaces' basis are linear independed that doesn't necessarily mean their union is also linear independent,
but i can draw this conclusion (that this two are linear independent and are basis for V) from $dim(V)=dim(U)+dim(W)$ , i don't need $W\cap U=\left \{ \vec0 \right \}$ for that.

so actually, my question is:
if $dim(V)=dim(U)+dim(W)$ , why doesn't it mean that U and W's basis are *disjoint sets?

*of course, disjoint except their common $\vec0$ , but that goes without saying, since U and W are subspaces.

**GJA** · January 6th 2013, 10:51 PM

Hi Stormey,

In the previous post you said

i'm aware that if two subspaces' basis are linear independed that doesn't necessarily mean their union is also linear independent,
but i can draw this conclusion (that this two are linear independent and are basis for V) from $dim(V)= dim(U)+dim(W)$

However, knowing $dim(V) = dim(U) + dim(W)$ does not imply the bases of $U$ and $W$ are linearly independent. For example, take $U=W=span([1,0])$ and $V=\mathbb{R}^{2}.$ Then $dim(V) = dim(U)+dim(W)$ holds, but the bases for $U$ and $W$ are not linearly independent.

Does this clear things up? Good luck!

**Stormey** · January 6th 2013, 11:04 PM

Brilliant, thanks man!

I forgot that U can be equal to W.
It all makes sense now.

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