Two Ways to Combine Subspaces
Suppose two subspaces \(U\) and \(W\) lie in the same vector space \(V\). Some vectors may belong to both; others may be made by adding a vector from \(U\) to a vector from \(W\). These two ideas produce the intersection and the sum. Their definitions look similar in that both combine information from \(U\) and \(W\), but their membership tests are different: intersection requires one vector to satisfy both conditions, while sum requires a suitable decomposition.
Throughout the general arguments, let \(V\) be a vector space over a field \(\mathbb{F}\), and let \(U,W\) be subspaces of \(V\). A subspace is a subset containing the zero vector and closed under addition and scalar multiplication. We use the subspace test: a nonempty subset \(S\) of \(V\) is a subspace if and only if \(a\mathbf{x}+b\mathbf{y}\in S\) whenever \(\mathbf{x},\mathbf{y}\in S\) and \(a,b\in\mathbb{F}\). Indeed, a subspace has this closure by its addition and scalar-multiplication properties. Conversely, if \(S\) is nonempty and has this closure, choose \(\mathbf{x}\in S\); taking \(a=b=0\) shows \(\mathbf{0}\in S\), and choosing coefficients \((1,1)\) or \((a,0)\) gives closure under addition or scalar multiplication.
The decomposition in the sum need not be unique. The definition asks only whether there are vectors \(\mathbf{u}\in U\) and \(\mathbf{w}\in W\) whose sum is the given vector. For the intersection, by contrast, the vector itself must belong to each set. These distinctions will guide both the proofs and the examples.
Why the Intersection Is a Subspace
Proof. Since \(U\) and \(W\) are subspaces, both contain \(\mathbf{0}\). Therefore \(\mathbf{0}\in U\cap W\), so the intersection is nonempty. Let \(\mathbf{x},\mathbf{y}\in U\cap W\) and \(a,b\in\mathbb{F}\). Membership in the intersection gives \(\mathbf{x},\mathbf{y}\in U\) and \(\mathbf{x},\mathbf{y}\in W\). As \(U\) is a subspace, \(a\mathbf{x}+b\mathbf{y}\in U\); as \(W\) is a subspace, \(a\mathbf{x}+b\mathbf{y}\in W\). Hence \(a\mathbf{x}+b\mathbf{y}\in U\cap W\). The subspace test proves the claim. \(\square\)
The proof follows the membership condition literally: to show that a vector is in the intersection, show that it is in each of the two subspaces. The same combination \(a\mathbf{x}+b\mathbf{y}\) stays in each because each set is already closed under linear combinations of its own vectors.
Worked Example: Intersecting Two Coordinate Planes
In \(\mathbb{R}^3\), let \(U=\{(x,y,z):z=0\}\) and \(W=\{(x,y,z):y=0\}\). These are subspaces: each consists of vectors satisfying a homogeneous linear equation, and linear combinations of vectors satisfying that equation still satisfy it. Determine the intersection and test three vectors.
A vector lies in \(U\cap W\) precisely when it satisfies both \(z=0\) and \(y=0\). Its \(x\)-coordinate is unrestricted, so
The vector \((3,0,0)\) belongs to both planes, hence lies in the intersection. The vector \((3,1,0)\) lies in \(U\), since its third coordinate is \(0\), but not in \(W\), since its second coordinate is \(1\); it is not in the intersection. The vector \((3,0,1)\) lies in \(W\) but not in \(U\), so it also fails the intersection condition.
Check: \((3,0,0)\) has both its second and third coordinates equal to \(0\). Each of the other two vectors violates exactly one of the two defining equations.
Why the Sum Is a Subspace
Proof. Since \(\mathbf{0}\in U\) and \(\mathbf{0}\in W\), we have \(\mathbf{0}=\mathbf{0}+\mathbf{0}\in U+W\), so \(U+W\) is nonempty. Now take \(\mathbf{x},\mathbf{y}\in U+W\). By the definition of the sum, there are \(\mathbf{u}_1,\mathbf{u}_2\in U\) and \(\mathbf{w}_1,\mathbf{w}_2\in W\) such that \(\mathbf{x}=\mathbf{u}_1+\mathbf{w}_1\) and \(\mathbf{y}=\mathbf{u}_2+\mathbf{w}_2\). For any \(a,b\in\mathbb{F}\), distributivity gives \(a\mathbf{x}+b\mathbf{y}=(a\mathbf{u}_1+b\mathbf{u}_2)+(a\mathbf{w}_1+b\mathbf{w}_2)\). Because \(U\) is a subspace, \(a\mathbf{u}_1+b\mathbf{u}_2\in U\); because \(W\) is a subspace, \(a\mathbf{w}_1+b\mathbf{w}_2\in W\). Their sum therefore belongs to \(U+W\), by its definition. The subspace test proves the claim. \(\square\)
The proof depends on splitting each of \(\mathbf{x}\) and \(\mathbf{y}\) into its \(U\)-part and \(W\)-part. It does not require the decompositions to be unique: any decompositions that establish membership are enough to carry out the closure argument.
Worked Example: A Sum of Two Lines in \(\mathbb{R}^2\)
Let \(U=\{c(1,1):c\in\mathbb{R}\}\) and \(W=\{d(1,-1):d\in\mathbb{R}\}\). To test whether an arbitrary \((a,b)\) belongs to \(U+W\), seek \(c,d\in\mathbb{R}\) such that
Equating coordinates gives \(c+d=a\) and \(c-d=b\). Adding these equations yields \(2c=a+b\), so \(c=(a+b)/2\). Subtracting the second equation from the first yields \(2d=a-b\), so \(d=(a-b)/2\). These scalars exist for every real \(a,b\). Thus every vector in \(\mathbb{R}^2\) lies in \(U+W\), and \(U+W=\mathbb{R}^2\).
For instance, for \((5,1)\), the coefficients are \(c=(5+1)/2=3\) and \(d=(5-1)/2=2\). Hence \((5,1)=3(1,1)+2(1,-1)=(3,3)+(2,-2)=(5,1)\).
Check: The two summands are in \(U\) and \(W\), respectively, and their coordinatewise sum is \((3+2,3-2)=(5,1)\). The formulas for \(c\) and \(d\) also verify membership for every \((a,b)\), not just this example.
Reading Membership Conditions in Coordinates
When subspaces are given by coordinate descriptions, intersection membership often becomes a list of simultaneous equations. Sum membership is different: it asks whether the coordinates can be expressed as the sum of one allowed vector from each subspace. Solving for such a decomposition can either produce a witness or show that no decomposition exists.
Worked Example: A Sum That Is Not All of \(\mathbb{R}^3\)
In \(\mathbb{R}^3\), let \(U=\{(x,0,0):x\in\mathbb{R}\}\) and \(W=\{(0,y,0):y\in\mathbb{R}\}\). A vector of \(U+W\) has the form
Therefore \(U+W=\{(x,y,0):x,y\in\mathbb{R}\}\). The vector \((2,-3,0)\) belongs to the sum because \((2,-3,0)=(2,0,0)+(0,-3,0)\). In contrast, \((2,-3,1)\) does not belong: every vector formed by adding an element of \(U\) and an element of \(W\) has third coordinate \(0\), while this vector has third coordinate \(1\).
The intersection has a different condition. A vector in \(U\) has only a possible first coordinate, while a vector in \(W\) has only a possible second coordinate. To belong to both, all coordinates must be zero, so \(U\cap W=\{\mathbf{0}\}\).
Check: The displayed decomposition uses one vector from each subspace and adds to \((2,-3,0)\). No such decomposition can produce a nonzero third coordinate, confirming the exclusion of \((2,-3,1)\). The only vector that has both forms \((x,0,0)\) and \((0,y,0)\) is \((0,0,0)\).
Common Pitfall: The Sum Is Not the Union
The sum \(U+W\) is not the union \(U\cup W\). A vector belongs to the union if it is in at least one of the subspaces; a vector belongs to the sum if it can be written as a sum of one vector from each. In particular, the sum includes combinations that need not lie in either subspace by itself.
For a concrete error, take the \(U\) and \(W\) from the previous example. Both \((1,0,0)\) and \((0,1,0)\) belong to \(U\cup W\). A mistaken closure check might conclude that their sum also belongs to the union. But
and \((1,1,0)\) is in neither axis: its second coordinate is nonzero, so it is not in \(U\), and its first coordinate is nonzero, so it is not in \(W\). Thus the union is not closed under addition and is not a subspace. The vector \((1,1,0)\) does belong to \(U+W\), which is exactly the distinction between the two constructions.
| Set | Membership condition | What to check |
|---|---|---|
| \(U\cap W\) | \(\mathbf{x}\in U\) and \(\mathbf{x}\in W\) | The same vector satisfies both conditions |
| \(U+W\) | \(\mathbf{x}=\mathbf{u}+\mathbf{w}\) for some \(\mathbf{u}\in U,\mathbf{w}\in W\) | Produce or rule out a decomposition |
| \(U\cup W\) | \(\mathbf{x}\in U\) or \(\mathbf{x}\in W\) | Membership in at least one set; closure can fail |
Choosing the Right Proof Structure
For an intersection, begin with arbitrary vectors already in the intersection. Translate that membership into membership in each subspace, use closure separately in each, and conclude that the resulting vector is in both. For a sum, begin with arbitrary vectors in the sum and write down their decompositions before applying closure. Keeping those decompositions visible prevents a common gap: it is not enough to say that the sum is closed without identifying its \(U\)-part and \(W\)-part.
These proofs also show what the definitions do not require. An intersection can contain only the zero vector, as in the last example, and it is still a subspace. A sum can equal all of \(V\), as with the two lines in \(\mathbb{R}^2\), or only a smaller subspace, as with the coordinate axes in \(\mathbb{R}^3\). Neither conclusion follows from the words “intersection” or “sum” alone; it follows from the specific subspaces involved.
Check Your Understanding
For each question, use the stated membership condition and give a reason for your conclusion.
- In \(\mathbb{R}^3\), let \(U=\{(x,y,z):x=0\}\) and \(W=\{(x,y,z):z=0\}\). Describe \(U\cap W\).
- For the subspaces in question 1, decide whether \((0,4,0)\) belongs to \(U\cap W\), to \(U+W\), or to both.
- Let \(U=\operatorname{span}\{(1,0)\}\) and \(W=\operatorname{span}\{(0,1)\}\) in \(\mathbb{R}^2\). Write \((3,-2)\) as a sum of one vector from each subspace.
- State the closure step needed to prove that \(U\cap W\) is a subspace, starting with arbitrary \(\mathbf{x},\mathbf{y}\in U\cap W\).
- Using the coordinate axes in \(\mathbb{R}^3\), give two vectors in \(U\cup W\) whose sum is not in \(U\cup W\).
