= Solution
A <vector space> over the <real numbers> is a <set> $V$ with an <addition> making it an <abelian group> and a <scalar multiplication> by $\mathbb R$ satisfying, for <scalars> $a,b$ and <vectors> $u,v$, the identities $a(u+v)=au+av$, $(a+b)v=av+bv$, $a(bv)=(ab)v$ and $1v=v$. A <vector subspace> is a nonempty subset closed under <addition> and <scalar multiplication>; a <proper vector subspace> is one unequal to $V$. A <spanning set> is a <set> whose finite <linear combinations> give every <vector> of $V$. A <basis> is a <linearly independent> <spanning set>. The <dimension> is the number of <vectors> in a <basis>, or its cardinality when the <basis> is infinite.
The <sum of vector subspaces> and their <intersection of vector subspaces> are
$$
U+W=\{u+w:u\in U,\ w\in W\},\qquad
U\cap W=\{v\in V:v\in U\text{ and }v\in W\}.
$$
Both are <vector subspaces>: <addition> and <scalar multiplication> preserve the displayed forms or simultaneous membership. The <intersection of vector subspaces> is never empty because \b[$0\in U\cap W$].
For the two given <hyperplanes>, adding and subtracting their defining equations gives $x_1=x_2$ and $x_3=x_4$. Hence
$$
U\cap W=\{(a,a,b,b):a,b\in\mathbb R\}
=\operatorname{span}\{b_1,b_2\},\quad b_1=(1,1,0,0),\quad b_2=(0,0,1,1).
$$
These two nonzero <vectors> are <orthogonal>, and consequently form an <orthogonal basis> of the <intersection of vector subspaces>. Put $u=(1,-1,-1,1)$ and $w=(1,-1,1,-1)$. Direct substitution gives $u\in U$, $w\in W$. Their <dot products> with $b_1,b_2$ vanish, and $u\cdot w=0$. Each defining equation of $U$ or $W$ is one nonzero linear constraint in four variables, so each <vector subspace> has <dimension> three. Thus each space has the indicated <orthogonal basis>:
$$
\boxed{U:\ (b_1,b_2,u),\qquad W:\ (b_1,b_2,w),\qquad U+W:\ (b_1,b_2,u,w).}
$$
The last four <vectors> are nonzero and mutually <orthogonal>, so they are a <basis> of $\mathbb R^4$. They lie in $U+W$, proving $U+W=V$. The <dimension> identity is therefore verified numerically:
$$
\boxed{\dim U+\dim W=3+3=4+2=\dim(U+W)+\dim(U\cap W).}
$$
Back to article page
