# Fairly basic question about direct sums and sums of vector spaces

Consider the following three subspaces of $$\mathbb{R}^3$$: $$U = [(a, a − b, a + b) : a, b ∈ R], V = [(0, 0, c) : c ∈ R], W = [(d, e, e) : d, e ∈ R]$$

(i) Find the (unique) $$u ∈ U$$ and $$v ∈ V$$ with $$(2, 3, 4) = u + v$$. Now show that $$U ⊕ V = \mathbb{R}^3$$.

(ii) Show that $$U + W = \mathbb{R}^3$$, but that the sum is not direct.

So with part (i) I have found the unique $$u$$ and $$v$$: $$u = (2,3,1), v= (0,0,3)$$, and am trying to figure out how to prove the direct sum is equal to $$\mathbb{R}^3$$. I know that the only intersection of the subspaces is the zero vector, but don't know how to show that it the direct sum spans $$\mathbb{R}^3$$.

For part (ii) I understand that there will be intersections of the two subspaces which are not the zero vector, which will make them not a direct sum, but do not know how to show that, nor how to show that their sum is $$\mathbb{R}^3$$. Any help would be appreciated. Thanks.

(i) We can show that any element $$(x,y,z)\in \mathbb R^3$$ can be written as the sum of two elements in $$U$$ and $$V$$. Any such element is of the form $$(a,a-b,a+b)+(0,0,c)=(a,a-b,a+b+c)$$ for some $$a,b,c\in \mathbb R$$. So, we can pick $$a=x$$, $$b=x-y$$ and $$c=-(2x-y)+z$$. Then $$(x,y,z)=(a,a-b,a+b+c)$$ and thus $$U+V=\mathbb R^3$$. You have already shown that $$U\cap V=\{(0,0,0)\}$$ (if $$(a,a-b,a+b)=(0,0,c)$$, then $$a=b=0$$ and hence $$c=0$$), so the sum is direct.

(ii) This can be done similarly to (i), but I'll illustrate a slightly different method. The sum of any $$u\in U$$ and $$w\in W$$ is of the form $$u+w=(a,a-b,a+b)+(d,e,e)=(a+d,a-b+e,a+b+e)=(f+d-e,f-b,f+b) ~~~~\text{where f=a+e.}$$ Then $$u+w=f(1,1,1)+d(1,0,0)+e(0,-1,1)$$ for some $$f,d,e\in\mathbb R$$. Therefore $$U+W=\mathrm{Span}(A)$$, where $$A=\{(1,1,1),(1,0,0),(0,-1,1)\}$$. We'll show that $$A$$ is linearly independent in $$\mathbb R^3$$. If $$f(1,1,1)+d(1,0,0)+e(0,-1,1)=0$$ for some $$f,d,e\in \mathbb R$$ then $$(i)~f+d=0,~~~~~(ii)~f-e=0,~~~~~(iii)~f+e=0.$$ Sum $$(ii)$$ and $$(iii)$$ to obtain $$f=0$$ and substract the same equations to obtain $$e=0$$. Replacing in $$(i)$$ we get $$d=0$$ and thus $$A$$ is a linearly independent set in $$\mathbb R^3$$. However, $$\dim_{\mathbb R} \mathbb R^3=3$$, so $$\mathbb R^3 = \mathrm{Span}(A)=U+W$$ since $$A$$ is a linearly independent subset of $$\mathbb R^3$$ with $$3$$ elements.

There are various ways to show the direct sum $$U\oplus V$$ spans $$\mathbb R^3$$. But it depends on what facts you know about $$\mathbb R^3$$ in particular and about vector spaces in general.

If you can find vectors $$u\in U$$ and $$v\in V$$ so that $$u+v = (1,0,0)$$, and similarly for $$(0,1,0)$$ and $$(0,0,1),$$ then you can construct a basis of $$\mathbb R^3$$ in $$U\oplus V$$.

Alternatively, if you know that the dimension of a direct sum of subspaces must be the sum of the dimensions of the two subspaces, you can show that the dimension of $$U\oplus V$$ is $$3$$. If you also know that a direct sum of subspaces is a subspace of the same vector space and that the only subspace of the same dimension as a vector space is the vector space itself, you then can show that $$U\oplus V = \mathbb R^3$$.

For (ii), again if you know the dimension of a direct sum of subspaces must be the sum of the dimensions of the two subspaces, then if you can just show that the sum of $$U$$ and $$W$$ is $$\mathbb R^3$$ you already know the sum is not direct. But you can also show it is not direct by exhibiting a non-zero vector (one vector is enough) that is in both $$U$$ and $$W$$.

To prove that $$U + W = \mathbb R^3$$ you could exhibit pairs of vectors that add to $$(1,0,0)$$, $$(0,1,0)$$, and $$(0,0,1),$$ respectively. Alternatively, you could exhibit a vector in $$W$$ that is not in $$U,$$ thereby showing that $$U + W$$ has a higher dimension than $$U,$$ therefore at least dimension $$3$$; and again some of the facts from before will lead to the conclusion that the sum can only be $$\mathbb R^3$$ itself.

The other answer presents additional methods and they are all good too.