When Equal Dimension Forces Equality
Dimension counts the number of vectors in a basis, but a count by itself does not tell us which vectors belong to a space. The key additional information is containment. If a subspace \(W\) sits inside a finite-dimensional vector space \(V\), and a basis of \(W\) already has as many vectors as a basis of \(V\), there is no room for \(V\) to contain any additional independent direction. In fact, \(W\) must be all of \(V\).
All vector spaces in the main result are over a common field \(\mathbb{F}\), which may be \(\mathbb{R}\) or \(\mathbb{C}\). A subspace \(W\) of \(V\) is a subset that contains the zero vector and is closed under vector addition and scalar multiplication. A basis is an independent list that spans the space; in a finite-dimensional space, its number of vectors is the dimension. We write \(W\subseteq V\) for containment, allowing equality, and \(W\subsetneq V\) for proper containment.
The theorem is an equality test only when the subspace condition is present. Its main direction says that a subspace cannot be both proper and have the full dimension of its ambient space. We will prove this by showing that a proper subspace would allow us to construct one more independent vector than the dimension of \(V\).
Why a Proper Subspace Has Smaller Dimension
We use the following standard consequence of the basis and dimension results: in an \(n\)-dimensional vector space, every independent list has at most \(n\) vectors. Here is the coordinate argument that explains the bound. Take a basis of the space and express each vector of the list in that basis. If the list has \(p>n\) vectors, its coordinate columns form an \(n\times p\) matrix. The homogeneous system with this matrix has a nonzero solution: after row reduction there are at most \(n\) pivot columns, so at least one variable is free. That nonzero solution gives a nontrivial linear combination of the original vectors equal to zero, contradicting independence.
Proof. Choose a basis \(\mathcal{B}\) of \(V\), which has \(n\) vectors. For a list \(\mathbf{v}_1,\dots,\mathbf{v}_p\) with \(p>n\), form the \(n\times p\) matrix whose \(j\)th column is the coordinate vector \([\mathbf{v}_j]_{\mathcal{B}}\). Its homogeneous system has a nonzero solution \((c_1,\dots,c_p)^{\top}\), because row reduction gives at most \(n\) pivot columns for \(p>n\) variables. Thus \(\sum_{j=1}^{p}c_j[\mathbf{v}_j]_{\mathcal{B}}=\mathbf{0}\). Coordinates with respect to a basis preserve vector addition and scalar multiplication, so this equation says \(\sum_{j=1}^{p}c_j\mathbf{v}_j=\mathbf{0}\). The coefficients are not all zero, and therefore the list is dependent. \(\square\)
Now suppose \(W\subsetneq V\). Choose a basis \(\mathbf{w}_1,\dots,\mathbf{w}_k\) of \(W\). Because containment is proper, some vector \(\mathbf{v}\) belongs to \(V\) but not to \(W\). The list \(\mathbf{w}_1,\dots,\mathbf{w}_k,\mathbf{v}\) is independent: if a relation had a nonzero coefficient on \(\mathbf{v}\), it would express \(\mathbf{v}\) as a combination of the basis vectors of \(W\), putting \(\mathbf{v}\) in \(W\). If its coefficient on \(\mathbf{v}\) were zero, independence of the basis of \(W\) would force all remaining coefficients to be zero. This extra independent vector is the reason a proper subspace must have strictly smaller dimension.
Proof. If \(W=V\), then the spaces have the same dimension by the definition of dimension. Conversely, suppose \(\dim W=\dim V=n\). If \(n=0\), a zero-dimensional vector space consists only of the zero vector, so \(W=V=\{\mathbf{0}\}\). Now suppose \(n>0\), and let \(\mathbf{w}_1,\dots,\mathbf{w}_n\) be a basis of \(W\). Assume, for contradiction, that \(W\ne V\). There is then a vector \(\mathbf{v}\in V\) with \(\mathbf{v}\notin W\). Consider a relation \(\sum_{i=1}^{n}a_i\mathbf{w}_i+b\mathbf{v}=\mathbf{0}\). If \(b\ne0\), rearranging gives \(\mathbf{v}=-\sum_{i=1}^{n}(a_i/b)\mathbf{w}_i\), which would put \(\mathbf{v}\) in \(W\), a contradiction. Hence \(b=0\), and independence of the basis of \(W\) forces \(a_1=\cdots=a_n=0\). Therefore \(\mathbf{w}_1,\dots,\mathbf{w}_n,\mathbf{v}\) is an independent list of \(n+1\) vectors in the \(n\)-dimensional space \(V\). This contradicts the lemma. So \(W=V\), proving both directions. \(\square\)
Proof. Since \(W\) is a subspace of the finite-dimensional space \(V\), it is finite-dimensional. The dimension of a subspace cannot exceed the dimension of its ambient space, as follows by applying the lemma to a basis of the subspace. Thus \(\dim W\leq\dim V\). If equality held, the equal-dimension theorem would give \(W=V\), contrary to proper containment. Therefore \(\dim W<\dim V\). \(\square\)
Worked Applications
Worked Example: Two Subspaces of a Plane
Let \(V=\{(x,y,z)\in\mathbb{R}^3:z=x+y\}\), and let \(W=\operatorname{span}\{(1,1,2),(1,-1,0)\}\). Show that \(W=V\).
Every vector in \(V\) has the form \((x,y,x+y)=x(1,0,1)+y(0,1,1)\). The two displayed vectors are independent: if \(a(1,0,1)+b(0,1,1)=\mathbf{0}\), the first and second coordinates give \(a=0\) and \(b=0\). They therefore form a basis of \(V\), so \(\dim V=2\).
Both generators of \(W\) belong to \(V\): their third coordinates are respectively \(1+1=2\) and \(1+(-1)=0\). Since \(V\) is a subspace, their span satisfies \(W\subseteq V\). To check the dimension of \(W\), suppose \(a(1,1,2)+b(1,-1,0)=\mathbf{0}\). The first two coordinates give \(a+b=0\) and \(a-b=0\). Adding these equations yields \(2a=0\), so \(a=0\), and then \(b=0\). Thus the two generators are independent and form a basis of \(W\), giving \(\dim W=2\).
Since \(W\subseteq V\) and \(\dim W=\dim V=2\), the theorem gives \(W=V\). We can check the conclusion directly as well. For any \((x,y,x+y)\in V\), set \(a=(x+y)/2\) and \(b=(x-y)/2\). Then \(a+b=x\), \(a-b=y\), and \(2a=x+y\), so
Thus every vector of \(V\) belongs to \(W\), as required.
Worked Example: A Plane That Is Proper in R³
Let \(W=\{(x,y,z)\in\mathbb{R}^3:x+y+z=0\}\). Find its dimension and use it to decide whether \(W\) equals \(\mathbb{R}^3\).
Solving the equation for \(z\) gives \(z=-x-y\), so every vector in \(W\) has the form \((x,y,-x-y)=x(1,0,-1)+y(0,1,-1)\). The two spanning vectors are independent: if \(a(1,0,-1)+b(0,1,-1)=\mathbf{0}\), the first coordinate gives \(a=0\), and the second gives \(b=0\). Hence they form a basis and \(\dim W=2\), whereas \(\dim\mathbb{R}^3=3\).
The dimensions differ, so \(W\ne\mathbb{R}^3\). A direct check confirms proper containment: \((1,0,0)\in\mathbb{R}^3\), but its coordinate sum is \(1\), not \(0\), so \((1,0,0)\notin W\). In the other direction, the parameterization checks out because \(x+y+(-x-y)=0\) for every choice of \(x,y\).
Worked Example: Equal-Dimension Polynomial Subspaces
In the real vector space \(V=P_2(\mathbb{R})\) of polynomials of degree at most \(2\), let \(W=\operatorname{span}\{1+x,\ x+x^2,\ 1+x^2\}\). Show that \(W=V\).
The standard basis \((1,x,x^2)\) gives \(\dim V=3\). To prove that the three generators are independent, suppose \(a(1+x)+b(x+x^2)+c(1+x^2)=0\). Comparing the constant, \(x\), and \(x^2\) coefficients gives, respectively, \(a+c=0\), \(a+b=0\), and \(b+c=0\). The second equation gives \(b=-a\); the third then gives \(c=a\); the first becomes \(2a=0\). Over \(\mathbb{R}\), this forces \(a=0\), followed by \(b=c=0\). Thus the generators form a basis of \(W\), and \(\dim W=3\).
By construction, \(W\subseteq V\). The dimensions agree, so \(W=V\). For a direct check, take any \(A+Bx+Cx^2\in V\), and define \(a=(A+B-C)/2\), \(b=(-A+B+C)/2\), and \(c=(A-B+C)/2\). The constant coefficient of \(a(1+x)+b(x+x^2)+c(1+x^2)\) is \(a+c=(A+B-C+A-B+C)/2=A\). Its \(x\) coefficient is \(a+b=(A+B-C-A+B+C)/2=B\). Its \(x^2\) coefficient is \(b+c=(-A+B+C+A-B+C)/2=C\). Therefore this combination is exactly \(A+Bx+Cx^2\), verifying that every polynomial in \(V\) lies in \(W\).
The Inclusion Condition and a Common Mistake
The phrase “same dimension” is not enough on its own. The theorem compares a subspace with the space containing it. Two different subspaces can have equal dimension when neither contains the other. For example, in \(\mathbb{R}^2\), let \(U=\operatorname{span}\{(1,0)\}\) and \(Z=\operatorname{span}\{(0,1)\}\). Each has dimension \(1\), but they are not equal: \((1,0)\in U\), while \((1,0)\notin Z\), since \(t(0,1)=(1,0)\) would require both \(t=0\) and \(1=0\).
Common mistake: “\(\dim U=\dim Z=1\), therefore \(U=Z\).” This calculation does not establish equality because it omits the required containment. The correction is to check the hypotheses: if one space is known to be a subspace of the other, equal dimensions do force equality; if there is no containment, equal dimensions alone do not settle the question. When working in an ambient space, first verify that the proposed smaller space really is a subspace, and only then compare dimensions.
The same reasoning can be useful even when equality is not the goal. If a proposed subspace is proper, its dimension must be strictly smaller than the ambient dimension. Consequently, finding equal dimensions is a quick way to prove equality once containment is established. Conversely, finding a vector of the ambient space outside the subspace proves proper containment and rules out equal dimensions.
Check Your Understanding
For each question, identify the containment hypothesis before using a dimension comparison.
- Let \(V\) be \(5\)-dimensional and let \(W\subseteq V\) be a subspace of dimension \(5\). What can be concluded, and which hypothesis makes the conclusion valid?
- Can a proper subspace of a \(4\)-dimensional space have dimension \(4\)? Justify your answer using the theorem.
- For \(W=\{(x,y,z)\in\mathbb{R}^3:x-y=0\}\), find a basis and compare \(\dim W\) with \(\dim\mathbb{R}^3\).
- Give two distinct one-dimensional subspaces of \(\mathbb{R}^2\), and explain why their equal dimensions do not imply equality.
- Explain why the proof of the equal-dimension theorem must account for the case \(\dim V=0\).
