The question reads:

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.


2 Answers 2


(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.


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