A Connected Strategy for Basis and Dimension Problems
Many basis-and-dimension problems begin with information in one form and ask for an answer in another. A subspace might be described by equations, by a list of generators, or by conditions on polynomials. A reliable solution identifies the subspace first, then finds a list that both spans it and is linearly independent. The length of that basis gives the dimension when the subspace is finite-dimensional.
Throughout, scalars are real. A linear combination of a list \((\mathbf{v}_1,\dots,\mathbf{v}_p)\) is a vector \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\), where \(c_1,\dots,c_p\in\mathbb{R}\). Its span is the set of all such combinations.
The two requirements for a basis do different jobs: spanning says that no vector of the subspace is missing, while independence says that the list has no redundant vector. In a multi-step problem, it is important to verify both. A list whose vectors all satisfy the equations defining \(W\) need not span \(W\); a list that spans \(W\) need not be independent.
From Equations to a Basis
For a subspace defined by homogeneous linear equations, solve the equations and express the general solution using free parameters. Each parameter gives one direction vector. The result is a spanning list; the independent free parameters also make it possible to establish that the list is a basis.
Proof. Let \(W\) be the set of solutions. The stated parametrization says that each \(\mathbf{x}\in W\) is a linear combination of \(\mathbf{u}_1,\dots,\mathbf{u}_k\), so \(W\subseteq\operatorname{span}\{\mathbf{u}_1,\dots,\mathbf{u}_k\}\). Each \(\mathbf{u}_j\) is itself a solution: set \(s_j=1\) and all other parameters to zero. Since the system is homogeneous, linear combinations of solutions are solutions, so the span is contained in \(W\). Thus the span equals \(W\). Now suppose \(c_1\mathbf{u}_1+\cdots+c_k\mathbf{u}_k=\mathbf{0}\). In the free coordinate corresponding to \(s_j\), the vector \(\mathbf{u}_j\) has entry \(1\), and every other \(\mathbf{u}_i\) has entry \(0\). The \(j\)-th free coordinate of the combination is therefore \(c_j\). Since the combination is zero, \(c_j=0\) for every \(j\). The list is independent and hence is a basis of \(W\). \(\square\)
Worked Example: A Subspace Given by Equations
Find a basis and dimension of the subspace \(W\subseteq\mathbb{R}^4\) defined by \(x_1+x_2-x_3=0\) and \(x_2+x_4=0\).
Use \(x_3=s\) and \(x_4=t\) as free variables. The second equation gives \(x_2=-t\). Substituting into the first gives \(x_1-t-s=0\), so \(x_1=s+t\). Thus every solution has the form
The proposition shows that these two direction vectors form a basis. Directly checking membership, the first vector gives \(1+0-1=0\) and \(0+0=0\); the second gives \(1-1-0=0\) and \(-1+1=0\). The parametrization also shows that every solution is in their span. Their free coordinates, \(x_3\) and \(x_4\), are respectively \((1,0)\) and \((0,1)\), so a combination equal to zero has both coefficients zero. Therefore \(\dim W=2\).
Check: Substituting the general vector gives \((s+t)+(-t)-s=0\) and \((-t)+t=0\), so every vector displayed satisfies both equations.
From Generators to a Basis
If a subspace is presented as the span of a list, place the vectors as columns of a matrix and row-reduce. Pivot positions identify which original vectors to keep. This method rests on two facts: elementary row operations preserve linear relations among columns, and the pivot columns of a reduced row echelon form are independent and span all its columns.
Proof. Row reduction is a sequence of elementary row operations, each represented by multiplication on the left by an invertible matrix. Their product \(\mathbf{E}\) is invertible, and \(\mathbf{R}=\mathbf{E}\mathbf{A}\). In \(\mathbf{R}\), each pivot column has a \(1\) in its pivot row and zeroes in the other rows. Every nonpivot column is a linear combination of these pivot columns: its entries in the pivot rows give the coefficients, and all its other entries are zero. Applying \(\mathbf{E}^{-1}\) to such a column equation shows that the corresponding original nonpivot column is a combination of the original pivot columns. Hence the original pivot columns span all original columns. If a combination of the original pivot columns is zero, applying \(\mathbf{E}\) gives the same combination of distinct pivot columns of \(\mathbf{R}\). Those columns have their \(1\)s in distinct rows, so each coefficient must be zero. The original pivot columns are independent as well as spanning, and thus form a basis. \(\square\)
Worked Example: Extracting a Basis from Generators
Find a basis and dimension for the subspace of \(\mathbb{R}^3\) spanned by \(\mathbf{v}_1=(1,0,1)\), \(\mathbf{v}_2=(0,1,1)\), \(\mathbf{v}_3=(1,1,2)\), and \(\mathbf{v}_4=(1,1,1)\).
Put the given vectors in the columns of a matrix. First use \(R_3\to R_3-R_1\), then \(R_3\to R_3-R_2\):
Scale row 3 by \(-1\), then use \(R_1\to R_1-R_3\) and \(R_2\to R_2-R_3\). The result is
The pivot positions are columns 1, 2, and 4. By the theorem, the corresponding original vectors \((\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_4)\) are a basis for the given span. In particular, the nonpivot column 3 is redundant. Indeed, \(\mathbf{v}_3=\mathbf{v}_1+\mathbf{v}_2\). To check independence of the selected vectors directly, suppose \(a\mathbf{v}_1+b\mathbf{v}_2+c\mathbf{v}_4=\mathbf{0}\). The coordinates give \(a+c=0\), \(b+c=0\), and \(a+b+c=0\). The first two equations imply \(a=b=-c\); the third then gives \(-c=0\), so \(a=b=c=0\). Thus the list is independent, spans the original list, and \(\dim W=3\).
Check: The omitted vector is recovered exactly: \((1,0,1)+(0,1,1)=(1,1,2)\). The three selected vectors are independent by the coordinate calculation above.
Extending a List in a Polynomial Subspace
The same reasoning applies in polynomial spaces. The elements of \(P_2\), the real vector space of polynomials of degree at most \(2\), have unique expressions \(a+bx+cx^2\). Conditions on polynomial values become equations in the coefficients. Once those equations are solved, one can test a proposed list for independence and add vectors that supply any missing directions.
Worked Example: A Basis for a Polynomial Subspace
Let \(W=\{p\in P_2:p(1)=0\}\). Starting with the independent list consisting of \(x-1\), extend it to a basis of \(W\) and find \(\dim W\).
Write \(p(x)=a+bx+cx^2\). The condition \(p(1)=0\) gives \(a+b+c=0\), so \(a=-b-c\). Therefore
This proves that \(W=\operatorname{span}\{x-1,x^2-1\}\): every polynomial in \(W\) has the displayed form, and each generator vanishes at \(1\). To check independence, suppose \(r(x-1)+s(x^2-1)\) is the zero polynomial. Expanding gives \(s x^2+(r)x-(r+s)=0\). Equality of polynomials requires equality of each coefficient, so \(s=0\) from the \(x^2\) coefficient and \(r=0\) from the \(x\) coefficient. Thus the two polynomials are a basis and \(\dim W=2\).
The starting list \((x-1)\) is independent because a scalar multiple \(r(x-1)\) is the zero polynomial only if its \(x\)-coefficient \(r\) is zero. Adding \(x^2-1\) extends it to the basis just verified.
Check: For example, \(p(x)=2+3x-5x^2\) satisfies \(p(1)=2+3-5=0\). Its coefficients \(b=3\), \(c=-5\) give \(3(x-1)-5(x^2-1)=2+3x-5x^2\), as required.
Common Pitfalls and a Practical Checklist
Do not use columns of the reduced matrix as vectors in the original subspace. For example, the single vector \((1,1)\) spans the line \(\operatorname{span}\{(1,1)\}\). Row reduction gives
The vector \((1,0)\) is not in the original line: if \((1,0)=c(1,1)\), the first coordinate requires \(c=1\), while the second requires \(c=0\). The reduced matrix helps locate pivot columns, but the basis vectors must be taken from the original matrix.
Another error is to show only that each proposed basis vector lies in \(W\). That proves containment of their span in \(W\), not equality. A complete span argument also shows that an arbitrary vector of \(W\) can be written as a combination of the proposed vectors. For equation-defined spaces, the general parametrization does this; for a generated space, express every original generator using the selected pivot columns.
For an integrated problem, proceed in a deliberate order:
- Write the subspace precisely, including both its defining conditions and its ambient space.
- If equations are given, solve them and use free variables to form direction vectors.
- If generators are given, place them as columns, row-reduce, and retain the corresponding original pivot columns.
- Verify spanning and independence, using the definition or the relevant parametrization or pivot argument.
- State the dimension only after a basis has been established; it is the number of vectors in that basis.
These steps keep the two defining features of a basis visible. They also make the dimension conclusion accountable: it comes from a verified basis, not just from counting equations, generators, or variables without checking what those counts mean.
Check Your Understanding
For each problem, identify the subspace first, then justify both spanning and independence for any proposed basis.
- Find a basis and dimension of the subspace of \(\mathbb{R}^3\) defined by \(x_1+x_2=0\).
- For \(\mathbf{a}_1=(1,1,0)\), \(\mathbf{a}_2=(0,1,1)\), and \(\mathbf{a}_3=(1,2,1)\), determine a basis for their span and its dimension.
- Find a basis and dimension of \(\{p\in P_2:p(0)=0\}\).
- Explain why the pivot columns of a reduced matrix indicate which columns to select, but are not generally the vectors to use as a basis of the original column span.
- Extend the independent list \((x)\) to a basis of \(P_2\).
