Why a Union Can Fail to Be a Subspace
A subspace is closed under addition: adding any two of its vectors must produce another vector in the subspace. If we take the union of two subspaces, each vector still comes from one of the original subspaces. But two vectors in the union might come from different subspaces, and their sum need not belong to either one. This is the obstruction that determines exactly when the union is a subspace.
Let \(V\) be a vector space over \(\mathbb{R}\). A subset \(U\) of \(V\) is a subspace if it contains the zero vector and is closed under addition and scalar multiplication: whenever \(\mathbf{u},\mathbf{v}\in U\) and \(c\in\mathbb{R}\), both \(\mathbf{u}+\mathbf{v}\in U\) and \(c\mathbf{u}\in U\). For two subsets \(U,W\subseteq V\), their union is \(U\cup W=\{\mathbf{v}\in V:\mathbf{v}\in U\text{ or }\mathbf{v}\in W\}\). The word “or” includes the possibility that a vector belongs to both.
Proof. Since \(U\) is a subspace, \(\mathbf{0}\in U\), so \(\mathbf{0}\in U\cup W\). Now take any \(\mathbf{v}\in U\cup W\) and any \(c\in\mathbb{R}\). By the definition of union, either \(\mathbf{v}\in U\) or \(\mathbf{v}\in W\). If \(\mathbf{v}\in U\), closure of \(U\) under scalar multiplication gives \(c\mathbf{v}\in U\), hence \(c\mathbf{v}\in U\cup W\). If \(\mathbf{v}\in W\), the same reasoning using \(W\) gives \(c\mathbf{v}\in U\cup W\). Thus the union contains zero and is closed under scalar multiplication. By the definition of subspace, it is a subspace if and only if it is also closed under addition. \(\square\)
The Containment Criterion
Proof. First suppose \(U\subseteq W\). Every element of \(U\) is then already in \(W\), so \(U\cup W=W\). Since \(W\) is a subspace, the union is a subspace. If instead \(W\subseteq U\), then \(U\cup W=U\), which is a subspace as well. This proves that either containment is sufficient.
For the converse, suppose \(U\cup W\) is a subspace. We show that the two subspaces must be comparable by containment. If neither were contained in the other, there would be vectors \(\mathbf{u}\in U\) with \(\mathbf{u}\notin W\), and \(\mathbf{w}\in W\) with \(\mathbf{w}\notin U\). Both vectors lie in \(U\cup W\), so closure under addition would give \(\mathbf{u}+\mathbf{w}\in U\cup W\). If \(\mathbf{u}+\mathbf{w}\in U\), then \(\mathbf{u}\in U\) and closure of \(U\) under subtraction imply \(\mathbf{w}=(\mathbf{u}+\mathbf{w})-\mathbf{u}\in U\), a contradiction. Here subtraction is permitted because \(-\mathbf{u}\in U\) by scalar closure and \(U\) is closed under addition. If \(\mathbf{u}+\mathbf{w}\in W\), then \(\mathbf{w}\in W\) and the same reasoning in \(W\) gives \(\mathbf{u}=(\mathbf{u}+\mathbf{w})-\mathbf{w}\in W\), again a contradiction. The sum cannot lie in either subspace, contradicting its membership in their union. Thus one of \(U\subseteq W\) or \(W\subseteq U\) must hold. \(\square\)
The theorem gives a useful way to test the union. If one subspace is contained in the other, the union is simply the larger subspace. If neither contains the other, choose one vector from each subspace that is not in the other. Their sum cannot be in either subspace, so addition fails. The examples below show how to carry out that test.
When One Subspace Is Contained in the Other
Worked Example: A Line Inside a Plane
In \(V=\mathbb{R}^3\), let \(U=\{(a,0,0):a\in\mathbb{R}\}\) and \(W=\{(a,b,0):a,b\in\mathbb{R}\}\). Both are subspaces: the zero vector is obtained by setting the parameters to zero, and adding their vectors or multiplying by a real scalar preserves the displayed coordinate form.
Every vector \((a,0,0)\in U\) also belongs to \(W\), by taking the second parameter \(b=0\). Thus \(U\subseteq W\), and \(U\cup W=W\). In particular, for arbitrary \((a,b,0),(c,d,0)\in W\) and \(t\in\mathbb{R}\),
Check: the union has exactly the same vectors as \(W\), so the displayed addition and scalar multiplication checks verify its closure. This is the nested case of the criterion.
When Addition Leaves the Union
Worked Example: The Two Coordinate Axes
In \(\mathbb{R}^2\), let \(U=\{(a,0):a\in\mathbb{R}\}\) be the horizontal axis and \(W=\{(0,b):b\in\mathbb{R}\}\) the vertical axis. Each is a subspace: its vectors have the indicated zero coordinate, which remains zero under addition and scalar multiplication.
Take \(\mathbf{u}=(1,0)\in U\) and \(\mathbf{w}=(0,1)\in W\). The vector \(\mathbf{u}\) is not in \(W\), and \(\mathbf{w}\) is not in \(U\), so neither subspace contains the other. Their sum is
The vector \((1,1)\) is not in \(U\), because its second coordinate is \(1\ne0\); it is not in \(W\), because its first coordinate is \(1\ne0\). Hence \((1,1)\notin U\cup W\), even though both summands are in the union.
Check: this explicitly violates closure under addition. The vectors chosen also demonstrate the two noncontainments required by the theorem.
Worked Example: Constant and Linear Polynomials
Let \(V=P_2(\mathbb{R})\), the vector space of real polynomials of degree at most \(2\), including the zero polynomial. Define \(U\) to be the constant polynomials and \(W=\{bx:b\in\mathbb{R}\}\), the polynomials that are scalar multiples of \(x\). Both are subspaces: sums and scalar multiples of constants remain constant, and sums \(b_1x+b_2x=(b_1+b_2)x\) and scalar multiples \(c(bx)=(cb)x\) remain in \(W\); both contain the zero polynomial.
The polynomial \(1\) belongs to \(U\) but not \(W\), and \(x\) belongs to \(W\) but not \(U\). Add them:
This is not constant, since its coefficient of \(x\) is \(1\). It is not a scalar multiple of \(x\), since its constant coefficient is \(1\). Therefore \(1+x\notin U\cup W\), although \(1\in U\cup W\) and \(x\in U\cup W\).
Check: comparing coefficients verifies nonmembership in each subspace: a constant polynomial has \(x\)-coefficient \(0\), while a multiple of \(x\) has constant coefficient \(0\). The sum has neither property.
Worked Example: Two Coordinate Planes
In \(\mathbb{R}^3\), set \(U=\{(a,b,0):a,b\in\mathbb{R}\}\) and \(W=\{(0,b,c):b,c\in\mathbb{R}\}\). The defining zero coordinate is preserved under addition and scalar multiplication, so each set is a subspace.
The vector \(\mathbf{u}=(1,0,0)\) is in \(U\) but not \(W\), because its first coordinate is nonzero. The vector \(\mathbf{w}=(0,0,1)\) is in \(W\) but not \(U\), because its third coordinate is nonzero. Their sum is \((1,0,1)\). Its third coordinate is nonzero, so it is not in \(U\); its first coordinate is nonzero, so it is not in \(W\).
Check: the addition calculation is \((1,0,0)+(0,0,1)=(1,0,1)\), and the coordinate tests show that the result is outside \(U\cup W\). Thus even though both sets are planes through the origin, their union is not a subspace.
Common Pitfall: Combining Membership with Closure
A tempting but incorrect argument is: “The first vector belongs to \(U\), the second belongs to \(W\), and both sets are subspaces, so their sum belongs to the union.” The closure property for \(U\) applies when both summands are in \(U\); the closure property for \(W\) applies when both are in \(W\). Neither property covers a pair with one vector in each subspace.
For a concrete wrong calculation, take the coordinate axes and claim that \((1,0)+(0,1)\) lies on one of the axes because each summand lies on an axis. The actual calculation gives \((1,1)\), whose two coordinates are both nonzero, so it lies on neither axis. The correction is to test the sum against each subspace’s defining condition, rather than infer membership from the summands.
Another useful distinction is that “union” does not mean “all sums of a vector from each subspace.” The union contains vectors already in \(U\) or already in \(W\). A sum such as \(\mathbf{u}+\mathbf{w}\) may be outside it. Whether such a sum is included in some other construction is a separate question; the union criterion depends only on containment.
Check Your Understanding
For each question, use the subspace conditions and the containment criterion where appropriate.
- Let \(U=\{(a,0):a\in\mathbb{R}\}\) and \(W=\{(a,b):a,b\in\mathbb{R}\}\). Is \(U\cup W\) a subspace? State the containment that decides.
- In \(\mathbb{R}^2\), choose one vector from each coordinate axis whose sum is outside the union, and verify the nonmembership.
- Let \(U\) be the constant polynomials and \(W=\{bx:b\in\mathbb{R}\}\) in \(P_2(\mathbb{R})\). Explain why \(1+x\) is not in \(U\cup W\).
- Suppose \(U,W\) are subspaces and \(U\cup W\) is a subspace. Prove that if \(U\not\subseteq W\), then \(W\subseteq U\).
- Why does closure of \(U\) under addition not by itself establish closure of \(U\cup W\) for a pair with one vector in \(U\) and the other in \(W\)?
