Subsets That Keep the Vector Space Operations
A subset of a vector space can itself have the structure of a vector space without changing how its elements are added or multiplied by scalars. The key phrase is inherited operations: for vectors in the subset, addition and scalar multiplication are exactly the operations already defined in the ambient space. In \(\mathbb{R}^n\), these are coordinatewise addition and scalar multiplication.
Throughout this tutorial the scalar field is \(\mathbb{R}\). The zero vector in \(\mathbb{R}^n\) is \(\mathbf{0}=(0,\dots,0)^\top\). A subset \(U\subseteq\mathbb{R}^n\) is a subspace if it is a vector space under the operations inherited from \(\mathbb{R}^n\). This means that its elements must remain in \(U\) when added or multiplied by real scalars, and that the vector space axioms hold for those operations.
When working with a particular subset, it is useful to separate two issues. First, check that the operations keep vectors inside the subset. Second, note that the algebraic identities for these operations already hold in \(\mathbb{R}^n\). For example, if \(\mathbf{u},\mathbf{v}\in U\), then \(\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}\) because this equality holds in \(\mathbb{R}^n\); if the sum also belongs to \(U\), the equality holds there as well. The examples below make the closure checks explicit.
The Zero Set and the Whole Space
Two examples are present in every vector space. The zero set here means the singleton set \(\{\mathbf{0}\}\), containing only the zero vector. The other example is the whole ambient space itself. Both use the original operations; neither requires a new definition of addition or scalar multiplication.
Proof. In \(\{\mathbf{0}\}\), the only possible pair of vectors to add is \(\mathbf{0}+\mathbf{0}=\mathbf{0}\), which remains in the set. For any \(c\in\mathbb{R}\), \(c\mathbf{0}=\mathbf{0}\), so scalar multiplication also remains in the set. The set contains the zero vector. The vector space identities inherited from \(V\) hold for these operations, so \(\{\mathbf{0}\}\) is a vector space. For the whole space \(V\), sums and scalar multiples of its elements are in \(V\) by the vector space axioms. It already satisfies all those axioms by assumption. Thus \(V\) is also a subspace of itself. \(\square\)
Worked Example: The Zero Set and the Whole of \(\mathbb{R}^2\)
Check the operations on \(\{\mathbf{0}\}\subseteq\mathbb{R}^2\). Write \(\mathbf{0}=(0,0)^\top\). The only possible sum is
For an arbitrary real scalar \(c\), multiplication gives
Both results are in \(\{\mathbf{0}\}\), so the operations remain in the set. The vector-space identities hold because these are the same operations as in \(\mathbb{R}^2\). Thus \(\{\mathbf{0}\}\) is a subspace. The set \(\mathbb{R}^2\) is also a subspace of itself: for arbitrary \(\mathbf{u}=(u_1,u_2)^\top\), \(\mathbf{v}=(v_1,v_2)^\top\), and \(c\in\mathbb{R}\), the vectors \((u_1+v_1,u_2+v_2)^\top\) and \((cu_1,cu_2)^\top\) are both in \(\mathbb{R}^2\).
Check: the zero set has not acquired any new operations: its sum and scalar multiple are still calculated in \(\mathbb{R}^2\), and the results remain \(\mathbf{0}\). The whole space includes every result of those operations by definition.
A Coordinate Plane Through the Origin
Consider the coordinate plane \(P\) in \(\mathbb{R}^3\) whose third coordinate is zero. Its vectors have the form \((x,y,0)^\top\), with \(x,y\in\mathbb{R}\). To establish that this set is a subspace, check that it contains the zero vector and that addition and scalar multiplication preserve the zero third coordinate.
Proof. Taking \(x=0\) and \(y=0\) shows that \(\mathbf{0}\in P\). Let \(\mathbf{u}=(x_1,x_2,0)^\top\) and \(\mathbf{v}=(y_1,y_2,0)^\top\) be arbitrary vectors in \(P\). Then \(\mathbf{u}+\mathbf{v}=(x_1+y_1,x_2+y_2,0)^\top\), which belongs to \(P\). For any \(c\in\mathbb{R}\), \(c\mathbf{u}=(cx_1,cx_2,0)^\top\), which also belongs to \(P\). Therefore the operations remain in \(P\), and the vector-space identities hold because they hold in \(\mathbb{R}^3\) for these inherited operations. Hence \(P\) is a subspace. \(\square\)
Worked Example: Adding and Scaling in a Coordinate Plane
Let \(\mathbf{u}=(1,-2,0)^\top\) and \(\mathbf{v}=(3,4,0)^\top\). Both vectors are in \(P\) because their third coordinates are zero. Add them coordinate by coordinate:
The resulting third coordinate is still zero, so the sum is in \(P\). Now multiply \(\mathbf{u}\) by \(-2\):
This vector is also in \(P\). These two calculations illustrate closure for particular vectors; the proposition establishes it for every pair of vectors in \(P\) and every real scalar.
Check: the sum has coordinates \((4,2,0)^\top\), and the scalar multiple has coordinates \((-2,4,0)^\top\). Each satisfies the defining coordinate condition that the third coordinate is zero.
A Line Through the Origin
Another simple coordinate-defined subset is the set of vectors in \(\mathbb{R}^3\) whose coordinates are in the ratio \(1:2:-1\). Equivalently, its elements are \((t,2t,-t)^\top\) for real \(t\). The common parameter matters: the coordinates cannot be chosen independently.
Worked Example: A Coordinate-Defined Line
Let \(L=\{(t,2t,-t)^\top:t\in\mathbb{R}\}\). First, \(t=0\) gives \((0,0,0)^\top\), so the zero vector is in \(L\). Take arbitrary elements \(\mathbf{u}=(a,2a,-a)^\top\) and \(\mathbf{v}=(b,2b,-b)^\top\) of \(L\). Their sum is
Since \(a+b\in\mathbb{R}\), the final vector is in \(L\). For an arbitrary \(c\in\mathbb{R}\), scalar multiplication gives
which is in \(L\) with parameter \(ca\). Thus the set contains zero and remains closed under both operations. The vector-space identities are inherited from \(\mathbb{R}^3\), so \(L\) is a subspace.
For a numerical check, choose \(a=1\) and \(b=-2\). Then \(\mathbf{u}=(1,2,-1)^\top\), \(\mathbf{v}=(-2,-4,2)^\top\), and
This is the vector in \(L\) with parameter \(-1\). Multiplying \(\mathbf{u}\) by \(3\) gives \((3,6,-3)^\top\), the vector with parameter \(3\).
Check: the sum has second coordinate \(-2=2(-1)\) and third coordinate \(1=-(-1)\). The scalar multiple has second coordinate \(6=2(3)\) and third coordinate \(-3=-(3)\). Both calculations satisfy the defining coordinate pattern.
When a Coordinate Restriction Fails
Not every coordinate-defined subset is a subspace. A common failure occurs when a coordinate is fixed at a nonzero value. For example, the set \(Q=\{(x,y,z)^\top\in\mathbb{R}^3:x=1\}\) is a translated plane, not a plane through the origin. The zero vector has first coordinate \(0\), so it is not in \(Q\). This alone rules out \(Q\) as a subspace.
Worked Example: A Translated Plane Is Not a Subspace
Consider \(\mathbf{u}=(1,0,0)^\top\in Q\). If \(Q\) were a subspace, multiplying \(\mathbf{u}\) by the scalar \(2\) would have to produce an element of \(Q\). Instead,
The first coordinate of this result is \(2\), not \(1\), so \(2\mathbf{u}\notin Q\). This is a concrete failure of closure under scalar multiplication. Also, the zero vector is not in \(Q\), since its first coordinate is \(0\). Either failure is enough to show that \(Q\) is not a subspace.
Check: \(\mathbf{u}\) satisfies \(x=1\), but \(2\mathbf{u}\) has \(x=2\). The claimed closure would contradict the defining condition, so the set is not closed under scalar multiplication.
A different kind of failure occurs when a coordinate condition permits separate pieces that do not combine into one piece. For instance, \(S=\{(x,y)^\top\in\mathbb{R}^2:xy=0\}\) consists of points on the two coordinate axes. The vectors \((1,0)^\top\) and \((0,1)^\top\) both belong to \(S\), but their sum \((1,1)^\top\) does not, because \(1\cdot1\ne0\). Therefore \(S\) is not a subspace.
Common Mistakes and a Useful Check
Mistake: checking only that individual vectors satisfy the condition. In the set \(S\) above, both \((1,0)^\top\) and \((0,1)^\top\) satisfy \(xy=0\). It would be wrong to conclude that their set is a subspace on that basis. A subspace must also contain the sum of any two of its vectors. The correction is to calculate the sum: \((1,0)^\top+(0,1)^\top=(1,1)^\top\), and \((1)(1)=1\ne0\). Closure fails.
Mistake: changing the operations. A subset is tested using the same addition and scalar multiplication as its ambient space. For example, the coordinate plane \(P\) is not given a special rule for adding its vectors; it uses coordinatewise addition in \(\mathbb{R}^3\). A new operation would be a different structure and would not establish that the subset is a subspace under the inherited operations.
For a coordinate-defined candidate, a careful direct check usually begins with the zero vector, then takes arbitrary elements of the set and applies addition and scalar multiplication. If one of these operations produces a vector outside the set, the subset is not a subspace. If the operations preserve the set, the vector-space identities themselves are inherited from the ambient space. The calculations must use arbitrary vectors to establish closure; checking one convenient pair is illustrative, not a proof for all pairs.
Check Your Understanding
For each set, decide whether it is a subspace under the inherited operations. Give a direct justification.
- Is \(\{(x,y,z)^\top\in\mathbb{R}^3:z=0\}\) a subspace? Check addition and scalar multiplication using arbitrary elements.
- Is \(\{(t,-t,3t)^\top:t\in\mathbb{R}\}\) a subspace? Identify the parameter of a sum and of a scalar multiple.
- Show that \(\{(x,y)^\top\in\mathbb{R}^2:x=4\}\) is not a subspace by testing the zero vector or a scalar multiple.
- For \(S=\{(x,y)^\top:xy=0\}\), find two vectors in \(S\) whose sum is not in \(S\).
- Explain why checking one numerical sum in a coordinate plane does not by itself prove that every sum stays in the plane.
