Membership in Every Subspace
When several subspaces impose different restrictions on a vector, their intersection consists of the vectors that satisfy every restriction at once. For example, a vector in the intersection of two planes must lie in the first plane and in the second plane. The family of subspaces need not be finite: the same idea applies when there are infinitely many restrictions. The central fact is that imposing all these conditions still produces a subspace.
Let \(V\) be a vector space over a field \(\mathbb{F}\), and let \(I\) be a nonempty index set. For each \(i\in I\), let \(U_i\) be a subspace of \(V\). The index set may be finite or infinite. Their intersection is the set of vectors that belong to every one of the subspaces.
The intersection is itself a set of vectors in the same ambient space \(V\). To test whether it is a subspace, use the subspace test: it must contain \(\mathbf{0}\) and be closed under linear combinations. The key is that each condition can be checked in every \(U_i\), one at a time.
The Arbitrary-Intersection Theorem
Proof. Write \(U=\bigcap_{i\in I}U_i\). Since each \(U_i\) is a subspace, it contains the zero vector. Therefore \(\mathbf{0}\in U_i\) for every \(i\in I\), which means \(\mathbf{0}\in U\). Now take any \(\mathbf{u},\mathbf{v}\in U\) and any scalars \(a,b\in\mathbb{F}\). The definition of intersection gives \(\mathbf{u},\mathbf{v}\in U_i\) for every \(i\in I\). Because each \(U_i\) is a subspace, it is closed under linear combinations, so \(a\mathbf{u}+b\mathbf{v}\in U_i\) for every \(i\in I\). Hence \(a\mathbf{u}+b\mathbf{v}\in U\). The zero-vector and closure conditions establish that \(U\) is a subspace of \(V\). \(\square\)
This proof does not try to list the members of the family or perform a separate argument for each possible size of \(I\). For any vector in the intersection, membership holds in every \(U_i\) by definition; closure in each of those subspaces then gives closure in the intersection.
The theorem assumes a common ambient vector space and that every member of the family is a subspace of it. It does not say that an arbitrary collection of subsets has a subspace as its intersection. Also, the index set in the theorem is nonempty. By the usual convention, the intersection of an empty family of subsets of \(V\) is \(V\) itself, which is a subspace, but that case has no individual \(U_i\) conditions to enforce.
Solving All Constraints at Once
In concrete problems, subspaces are often described by homogeneous equations. A vector belongs to their intersection exactly when it satisfies every equation from every subspace description. Solve the combined system, rather than solving each condition and then treating the separate solution sets as alternatives.
Worked Example: Intersecting Two Planes in \(\mathbb{R}^3\)
Let \(U_1=\{(x,y,z)^{\top}\in\mathbb{R}^3:x+y+z=0\}\) and \(U_2=\{(x,y,z)^{\top}\in\mathbb{R}^3:x-y=0\}\). Find \(U_1\cap U_2\).
A vector in the intersection must satisfy both equations. From \(x-y=0\), we get \(y=x\). Substitute this into \(x+y+z=0\): \(x+x+z=0\), so \(z=-2x\). Let \(x=t\), where \(t\in\mathbb{R}\). Then \(y=t\) and \(z=-2t\), giving
To check the result, take any vector in the displayed set. Its first defining expression is \(t+t-2t=0\), and its second is \(t-t=0\), so it belongs to both planes. Conversely, the two equations forced \(y=t\) and \(z=-2t\), so every vector in the intersection is in the displayed set. The set contains \(\mathbf{0}\) at \(t=0\); and a linear combination with scalars \(a,b\in\mathbb{R}\) of the vectors for parameters \(t_1,t_2\) is the vector for parameter \(at_1+bt_2\). This also directly verifies that the answer is a subspace.
The computation combines the two conditions into a single system. This is the general method for homogeneous constraints: write down all equations that must hold, solve them simultaneously, and check that the resulting description satisfies every original condition.
Worked Example: Three Conditions on Quadratic Polynomials
In the vector space \(P_2(\mathbb{R})\) of real polynomials of degree at most \(2\), consider the three subspaces defined by \(p(0)=0\), \(p(1)=0\), and \(p(-1)=0\), respectively. Find their intersection.
Write an arbitrary polynomial in the ambient space as \(p(x)=a+bx+cx^2\), with \(a,b,c\in\mathbb{R}\). Enforcing the first condition gives \(p(0)=a=0\). The second gives \(p(1)=a+b+c=0\), which, using \(a=0\), becomes \(b+c=0\). The third gives \(p(-1)=a-b+c=0\), which becomes \(-b+c=0\). Thus \(c=-b\) from the second equation and \(c=b\) from the third. Together these imply \(b=-b\), so \(2b=0\); over \(\mathbb{R}\), this gives \(b=0\), and then \(c=0\). Also \(a=0\). Therefore
Check the result by substitution: the zero polynomial has value \(0\) at \(0\), \(1\), and \(-1\), so it satisfies all three constraints. The equations above show that any polynomial satisfying all three must have \(a=b=c=0\). Each condition is homogeneous in the coefficients, and enforcing all three at once leaves only the zero polynomial.
Infinite Families of Constraints
The same reasoning applies when there is no finite list of equations to finish solving. A sequence has one coordinate for each positive integer, and membership in an intersection may require a condition on every coordinate.
Worked Example: An Intersection in Sequence Space
Let \(\mathbb{R}^{\mathbb{N}}\) be the vector space of all real sequences, with addition and scalar multiplication defined coordinatewise. For each \(n\in\mathbb{N}\), set \(U_n=\{\mathbf{x}\in\mathbb{R}^{\mathbb{N}}:x_n=0\}\), where \(x_n\) is the \(n\)th coordinate of \(\mathbf{x}\). Find \(\displaystyle\bigcap_{n\in\mathbb{N}}U_n\).
First, each \(U_n\) is a subspace. The zero sequence has \(n\)th coordinate \(0\), so it belongs to \(U_n\). If \(\mathbf{x},\mathbf{y}\in U_n\) and \(a,b\in\mathbb{R}\), then the \(n\)th coordinate of their linear combination is \(ax_n+by_n=a\cdot0+b\cdot0=0\). Thus \(a\mathbf{x}+b\mathbf{y}\in U_n\).
A sequence in the intersection must satisfy \(x_n=0\) for every \(n\in\mathbb{N}\). Since every coordinate is zero, it is the zero sequence. Conversely, the zero sequence belongs to every \(U_n\). Hence
The check uses both directions: the coordinate condition forces every member of the intersection to be \(\mathbf{0}\), and \(\mathbf{0}\) satisfies every coordinate condition. The family is infinite, but the argument does not require a last index. It uses the same “for every \(n\)” condition that defines the intersection.
The Intersection as the Largest Common Subspace
There is another useful way to understand the intersection. It contains exactly the vectors shared by all the subspaces, so it is contained in each one. Moreover, any subspace that is contained in every \(U_i\) must itself be contained in their intersection.
Proof. Suppose \(W\subseteq U_i\) for every \(i\in I\). Take any \(\mathbf{w}\in W\). Then \(\mathbf{w}\in U_i\) for every \(i\), so the definition of intersection gives \(\mathbf{w}\in\bigcap_{i\in I}U_i\). Thus \(W\subseteq\bigcap_{i\in I}U_i\). Conversely, suppose \(W\subseteq\bigcap_{i\in I}U_i\). Each vector in the intersection belongs to every \(U_i\), so every vector of \(W\) belongs to every \(U_i\). Therefore \(W\subseteq U_i\) for every \(i\in I\). This proves both directions. \(\square\)
Taking \(W=\bigcap_{i\in I}U_i\) in the defining condition shows that the intersection is contained in every member of the family. The proposition explains its shared-content role: it is the largest subset, and in particular the largest subspace, contained in all of them. This is useful when a problem asks for the vectors satisfying several subspace conditions, rather than merely asking whether the result is a subspace.
Common Pitfall: Confusing “And” with “Or”
The intersection condition is “belongs to every subspace,” not “belongs to at least one.” For a concrete check, let \(U=\{(x,0)^{\top}:x\in\mathbb{R}\}\) be the horizontal axis in \(\mathbb{R}^2\) and \(W=\{(0,y)^{\top}:y\in\mathbb{R}\}\) be the vertical axis. A mistaken calculation might use “or” and claim that \((1,0)^{\top}\) belongs to \(U\cap W\), because it belongs to \(U\). But it does not belong to \(W\), since its second coordinate is \(0\) and its first is not \(0\). To be in the intersection, a vector must have the form \((x,0)^{\top}\) and also \((0,y)^{\top}\). Equality of these coordinates forces \(x=0\) and \(y=0\), so the correct intersection is \(\{\mathbf{0}\}\).
The reverse mistake is to think that separate solutions must be combined by taking their union. In this example, the union contains both axes, whereas the intersection retains only their common vector. When solving equations, write every condition together before drawing a conclusion about membership.
Check Your Understanding
For each question, use the requirement that a vector in an intersection must satisfy every defining condition.
- In \(\mathbb{R}^3\), find the intersection defined by \(x+y=0\) and \(y+z=0\).
- Explain why the intersection of any finite nonempty collection of subspaces is a subspace, using the arbitrary-intersection theorem.
- For \(U_n=\{\mathbf{x}\in\mathbb{R}^{\mathbb{N}}:x_n=1\}\), explain why \(U_n\) is not a subspace.
- In \(P_2(\mathbb{R})\), find all \(p(x)=a+bx+cx^2\) satisfying \(p(0)=0\) and \(p(1)=0\).
- State both directions of the criterion for a subset \(W\) to be contained in \(\bigcap_{i\in I}U_i\).
