How Large Can a Subspace Be?
A subspace is itself a vector space, so it makes sense to ask for its dimension. If the ambient space is finite-dimensional, however, it is not automatic from the definition that a subspace has a finite basis: its vectors might form an infinite set. The key is that a subspace cannot contain an independent list longer than one in the ambient space. This bound lets us build a basis for the subspace in finitely many steps.
Throughout, let \(\mathbb{F}\) denote either \(\mathbb{R}\) or \(\mathbb{C}\). A basis of a vector space is an ordered list that is linearly independent and spans the space. A vector space is finite-dimensional if it has a finite basis. Its dimension is the number of vectors in a basis; the invariance of this number is the result established in the earlier material on invariance of dimension. The zero vector space has the empty list as a basis, so its dimension is \(0\).
The defining properties of a subspace ensure that it is a vector space in its own right: it contains \(\mathbf{0}\) and is closed under linear combinations. In particular, a basis of \(W\) must consist of vectors from \(W\), not merely vectors from the larger space \(V\). We first record the size restriction on independent lists in a finite-dimensional space.
Independent Lists Cannot Exceed the Ambient Dimension
Proof. Let \(\mathcal{B}\) be a basis of \(V\) with \(n\) vectors, and suppose \(\mathbf{v}_1,\dots,\mathbf{v}_k\) is an independent list in \(V\). Write each \(\mathbf{v}_j\) in its unique coordinates relative to \(\mathcal{B}\), and place those coordinate vectors as the columns of an \(n\times k\) matrix \(\mathbf{A}\). A relation among the columns of \(\mathbf{A}\) is exactly a relation with the same coefficients among \(\mathbf{v}_1,\dots,\mathbf{v}_k\), by uniqueness of coordinates. Thus the columns of \(\mathbf{A}\) are independent. If \(k>n\), row reduction of \(\mathbf{A}\) can have at most \(n\) pivot columns, so at least one of its \(k\) columns has no pivot. The homogeneous system \(\mathbf{A}\mathbf{c}=\mathbf{0}\) then has a free variable: set one free variable to \(1\), the other free variables to \(0\), and solve for the pivot variables. This gives a nonzero solution, contradicting independence of the columns. Therefore \(k\le n\). \(\square\)
The lemma applies to lists, including lists with repeated vectors. It is not a claim that a set must have at most \(n\) elements: a finite-dimensional space can contain infinitely many vectors. Rather, it says that no list of more than \(n\) vectors in that space can be linearly independent.
The Dimension Theorem for Subspaces
Proof. Write \(\dim V=n\). Begin with the empty list, which is independent and spans the zero subspace. If the current list spans \(W\), stop. If it does not span \(W\), choose a vector \(\mathbf{w}\in W\) outside the span of the current list and append it.
Appending \(\mathbf{w}\) preserves independence. Indeed, suppose a linear combination of the enlarged list is zero. If the coefficient of \(\mathbf{w}\) were nonzero, we could solve for \(\mathbf{w}\) as a linear combination of the previous vectors, contrary to its choice. Therefore that coefficient is zero, and independence of the previous list forces all the other coefficients to be zero.
Every vector selected lies in \(W\), so each list produced is also an independent list in \(V\). By the lemma, such a list has at most \(n\) vectors. Consequently the process must stop: if it had reached \(n\) vectors without spanning \(W\), the rule would allow another choice and produce an independent list of \(n+1\) vectors in \(V\), contradicting the lemma. When the process stops, the list is both independent and spanning in \(W\), hence is a finite basis of \(W\). Its length is at most \(n\), so \(\dim W\le n=\dim V\). Dimensions are nonnegative, including the empty basis case, and therefore \(0\le\dim W\le\dim V\). \(\square\)
This argument also covers \(n=0\). In that case \(V=\{\mathbf{0}\}\), so its only subspace is \(W=\{\mathbf{0}\}\); the empty list is a basis and \(\dim W=0\). The inequality is then \(0\le0\le0\).
The proof has two distinct jobs. The independent-list bound guarantees that the selection process cannot continue indefinitely; stopping guarantees that the resulting list spans all of \(W\). A finite independent list in \(W\) alone would not be enough to establish that \(W\) is finite-dimensional.
Worked Examples
Worked Example: A Plane in \(\mathbb{R}^3\)
Let \[ W=\{(x,y,z)\in\mathbb{R}^3:x+2y-z=0\}. \] Find a basis of \(W\) and compare its dimension with that of \(\mathbb{R}^3\).
The equation gives \(z=x+2y\), so every vector in \(W\) has the form \[ (x,y,z)=(x,y,x+2y)=x(1,0,1)+y(0,1,2). \] Both displayed vectors belong to \(W\): their coordinates satisfy \(1+2(0)-1=0\) and \(0+2(1)-2=0\). Thus they span \(W\). To check independence, suppose \[ a(1,0,1)+b(0,1,2)=(0,0,0). \] The first coordinate gives \(a=0\), and the second gives \(b=0\). They are independent, so they form a basis of \(W\), and \(\dim W=2\). Since the standard basis gives \(\dim\mathbb{R}^3=3\), the comparison is \(2\le3\).
Check: The vector \((3,-1,1)\) satisfies \(3+2(-1)-1=0\). Its basis coordinates are \(3\) and \(-1\), and \[ 3(1,0,1)-(0,1,2)=(3,-1,1). \] So the proposed basis represents this vector and the defining equation is satisfied.
Worked Example: A Subspace of the Polynomial Space \(P_2\)
Let \(P_2\) be the real polynomials of degree at most \(2\), and let \(W=\{p\in P_2:p(1)=0\}\). Find a basis and compare dimensions.
Write \(p(x)=a+bx+cx^2\). The condition \(p(1)=0\) says \(a+b+c=0\), or \(a=-b-c\). Therefore \[ p(x)=b(x-1)+c(x^2-1). \] Every element of \(W\) is a combination of \(x-1\) and \(x^2-1\), and each of those polynomials vanishes at \(1\). To check independence, suppose \(r(x-1)+s(x^2-1)\) is the zero polynomial. Its coefficient of \(x^2\) is \(s\), so \(s=0\); its coefficient of \(x\) is then \(r\), so \(r=0\). Hence these polynomials are a basis and \(\dim W=2\). The monomial basis \((1,x,x^2)\) gives \(\dim P_2=3\), so \(2\le3\).
Check: Take \(b=2\) and \(c=-1\). Then \(a=-b-c=-1\), and \[ 2(x-1)-(x^2-1)=-1+2x-x^2. \] Its value at \(1\) is \(-1+2-1=0\), and its coefficients in the proposed basis are exactly \(2\) and \(-1\).
Worked Example: A Subspace Defined by Two Generators
In \(\mathbb{R}^4\), let \(W=\operatorname{span}\{\mathbf{u},\mathbf{v}\}\), where \[ \mathbf{u}=(1,2,0,1),\qquad \mathbf{v}=(0,1,1,0). \] Determine \(\dim W\).
By definition, \(\mathbf{u},\mathbf{v}\) span \(W\). To verify that they are independent, suppose \(a\mathbf{u}+b\mathbf{v}=\mathbf{0}\). The first coordinate gives \(a=0\); the third coordinate gives \(b=0\). Thus the two generators are independent and form a basis, so \(\dim W=2\). The standard basis of \(\mathbb{R}^4\) has four vectors, giving \(2\le4\).
Check: The combination \(3\mathbf{u}-2\mathbf{v}\) is \[ (3,6,0,3)-(0,2,2,0)=(3,4,-2,3). \] This vector is in \(W\), and the calculation recovers it from the stated basis with coefficients \(3\) and \(-2\).
Worked Example: The Zero Subspace
For any finite-dimensional \(V\), consider \(W=\{\mathbf{0}\}\). Its basis is the empty list: the empty list spans \(\{\mathbf{0}\}\), and it has no nontrivial linear relation. Therefore \(\dim W=0\), which satisfies \(0\le\dim V\).
Check: In \(\mathbb{R}^2\), the empty list spans the zero vector, and \(\dim\mathbb{R}^2=2\). Thus \(0=\dim W\le2\), as the theorem predicts.
What the Bound Does—and Does Not—Say
The theorem says every subspace has some basis with no more vectors than a basis of the ambient space. It does not say that every list of vectors in the subspace is a basis. For instance, in the plane from the first example, the three vectors \[ \mathbf{v}_1=(1,0,1),\qquad \mathbf{v}_2=(0,1,2),\qquad \mathbf{v}_3=(1,1,3) \] all satisfy \(x+2y-z=0\), but the list is dependent.
A common mistake is to argue, “all three vectors lie in the plane, so they form a basis of it.” This confuses membership with independence. The incorrect basis claim fails because \[ \mathbf{v}_1+\mathbf{v}_2-\mathbf{v}_3 =(1+0-1,\;0+1-1,\;1+2-3) =(0,0,0). \] The correction is that \(\mathbf{v}_3=\mathbf{v}_1+\mathbf{v}_2\), so the three-vector list is redundant. The first two vectors are independent and span the plane, and therefore form a basis of dimension \(2\).
Another distinction matters: a subspace can have infinitely many elements while still having finite dimension. The plane in \(\mathbb{R}^3\) contains infinitely many vectors, but its basis has only two vectors. Dimension counts vectors in a basis, not all the vectors contained in the space.
Subspace dimension is useful because it measures how many independent directions are available inside the subspace. The theorem guarantees that this count is finite whenever the ambient space is finite-dimensional, and bounds that count without requiring a particular description of the subspace. The zero subspace is the limiting case with no directions; other subspaces may have any dimension allowed by the ambient bound.
Check Your Understanding
Use the definition of a basis and the dimension theorem to justify each answer.
- If \(\dim V=5\) and \(W\) is a subspace, what are the possible values of \(\dim W\)?
- Why does an independent list of six vectors in a subspace of a \(4\)-dimensional space have to be dependent?
- What is the dimension of the zero subspace, and what list is its basis?
- In the first worked example, verify directly that \((2,1,4)\) belongs to \(W\) and give its coordinates in the displayed basis.
- Explain why three vectors in a \(2\)-dimensional subspace cannot themselves form a basis.
