# Thread: proving that two subspaces are direct sum

1. ## proving that two subspaces are direct sum

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...

2. ## Re: proving that two subspaces are direct sum

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!

3. ## Re: proving that two subspaces are direct sum

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.

4. ## Re: proving that two subspaces are direct sum

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!

5. ## Re: proving that two subspaces are direct sum

Brilliant, thanks man!

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