Two Separate Claims Make a Basis Proof
A proposed list can fail to be a basis in either of two ways: it may not produce every vector in the space, or it may contain redundancy. A complete proof therefore has two distinct jobs. First, show that an arbitrary vector in the space can be written using the list. Second, show that no nontrivial choice of coefficients combines the list to give the zero vector. These are separate claims; establishing one does not establish the other.
Throughout, spaces and scalars are over \(\mathbb{R}\). A finite list \((\mathbf{v}_1,\dots,\mathbf{v}_p)\) in a vector space \(V\) has span \(\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_p\}\), the set of all vectors \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\) for real scalars \(c_1,\dots,c_p\). The list spans \(V\) if this span equals \(V\). It is linearly independent if the equation \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\) forces every coefficient to be zero.
In a spanning proof, begin with an arbitrary vector \(\mathbf{x}\in V\), not just a convenient example, and find coefficients that express it in terms of the candidate vectors. In an independence proof, begin with the equation that a linear combination equals \(\mathbf{0}\), then show that its only possible coefficients are all zero. Coefficients found for spanning may depend on \(\mathbf{x}\); independence concerns every possible coefficient choice in the homogeneous equation.
Why Unique Coefficients Capture Independence
Proof. First suppose the list is linearly independent. If \(\mathbf{x}\) has two representations, then \(\mathbf{x}=c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=d_1\mathbf{v}_1+\cdots+d_p\mathbf{v}_p\). Subtracting the right-hand combinations gives \((c_1-d_1)\mathbf{v}_1+\cdots+(c_p-d_p)\mathbf{v}_p=\mathbf{0}\). Independence forces \(c_i-d_i=0\) for every \(i\), so the two representations have identical coefficients. Conversely, suppose every vector in the span has at most one representation. If \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\), then \(\mathbf{0}\) has this representation and also the representation \(0\mathbf{v}_1+\cdots+0\mathbf{v}_p\). By uniqueness, \(c_i=0\) for every \(i\). Thus the list is independent. \(\square\)
This theorem explains a useful proof pattern: to prove a list is a basis, show that every vector has a representation and that this representation is unique. Existence establishes spanning; uniqueness establishes independence. In computations, the second part is often handled directly by setting a linear combination equal to zero and solving the resulting coefficient equations.
Worked Example: A Candidate List in \(\mathbb{R}^3\)
Prove that \(\mathbf{v}_1=(1,1,0)\), \(\mathbf{v}_2=(1,0,1)\), and \(\mathbf{v}_3=(0,1,1)\) form a basis of \(\mathbb{R}^3\).
Spanning. Let \(\mathbf{x}=(r,s,t)\) be arbitrary in \(\mathbb{R}^3\). We seek scalars \(a,b,c\) such that \(a\mathbf{v}_1+b\mathbf{v}_2+c\mathbf{v}_3=\mathbf{x}\). Equating coordinates gives
Add the first two equations and subtract the third to obtain \(2a=r+s-t\). Similarly, adding the first and third and subtracting the second gives \(2b=r+t-s\); adding the second and third and subtracting the first gives \(2c=s+t-r\). Thus choose
Substitution verifies the equations: \(a+b=(2r)/2=r\), \(a+c=(2s)/2=s\), and \(b+c=(2t)/2=t\). Since \(\mathbf{x}\) was arbitrary, the list spans \(\mathbb{R}^3\).
Independence. Suppose \(a\mathbf{v}_1+b\mathbf{v}_2+c\mathbf{v}_3=\mathbf{0}\). The coordinate equations are \(a+b=0\), \(a+c=0\), and \(b+c=0\). Adding the first two and subtracting the third gives \(2a=0\), so \(a=0\). The first equation then gives \(b=0\), and the second gives \(c=0\). The only solution is the trivial one, so the list is independent. It is therefore a basis.
Check. For instance, \((2,3,4)=\frac12\mathbf{v}_1+\frac32\mathbf{v}_2+\frac52\mathbf{v}_3\). Indeed, the coordinates of this combination are \(\frac12+\frac32=2\), \(\frac12+\frac52=3\), and \(\frac32+\frac52=4\).
Coefficient Equations in Coordinate Spaces
When candidate vectors in \(\mathbb{R}^m\) are written as columns of a matrix \(\mathbf{A}\), a coefficient vector \(\mathbf{c}=(c_1,\dots,c_p)^\top\) satisfies \(\mathbf{A}\mathbf{c}=c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\). This gives a direct translation between the basis questions and systems of equations.
Proof. Suppose first that the columns form a basis. Since they span \(\mathbb{R}^m\), every \(\mathbf{b}\in\mathbb{R}^m\) has at least one expression \(\mathbf{b}=c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\), which is a solution of \(\mathbf{A}\mathbf{c}=\mathbf{b}\). If \(\mathbf{c}\) and \(\mathbf{d}\) are two solutions, then \(\mathbf{A}(\mathbf{c}-\mathbf{d})=\mathbf{0}\). This is a linear combination of the columns equal to zero, so their independence gives \(\mathbf{c}-\mathbf{d}=\mathbf{0}\). Hence the solution is unique. Conversely, suppose \(\mathbf{A}\mathbf{c}=\mathbf{b}\) has exactly one solution for every \(\mathbf{b}\). Existence for every \(\mathbf{b}\) says that the columns span \(\mathbb{R}^m\). For independence, take any \(\mathbf{c}\) with \(\mathbf{A}\mathbf{c}=\mathbf{0}\). The zero vector also has the solution \(\mathbf{c}=\mathbf{0}\). Uniqueness for \(\mathbf{b}=\mathbf{0}\) forces the original \(\mathbf{c}\) to be zero. Thus the columns are independent and form a basis. \(\square\)
The theorem is a way to organize the proof, not a reason to skip its two parts. For an abstract space or a polynomial space, write the vector equation and compare its coefficients in the relevant representation. For a subspace, first take an arbitrary vector satisfying the subspace condition; this ensures the spanning argument applies to every vector in the space being considered.
Worked Example: A Candidate Basis for the Polynomial Space \(P_2(\mathbb{R})\)
Here \(P_2(\mathbb{R})\) is the space of real polynomials of degree at most two, including the zero polynomial. Prove that \(\mathbf{q}_1=1+x\), \(\mathbf{q}_2=x+x^2\), and \(\mathbf{q}_3=1+x^2\) form a basis.
Spanning. Take an arbitrary \(\mathbf{p}=c+bx+ax^2\) in \(P_2(\mathbb{R})\). Seek scalars \(u,v,w\) such that \(u\mathbf{q}_1+v\mathbf{q}_2+w\mathbf{q}_3=\mathbf{p}\). Comparing the constant, \(x\), and \(x^2\) coefficients gives
Adding the first two equations and subtracting the third gives \(2u=b+c-a\). Adding the second and third and subtracting the first gives \(2v=a+b-c\). Adding the first and third and subtracting the second gives \(2w=a+c-b\). Choose
These values satisfy the coefficient equations: \(u+w=c\), \(u+v=b\), and \(v+w=a\), by adding the displayed fractions. Hence \(u\mathbf{q}_1+v\mathbf{q}_2+w\mathbf{q}_3=\mathbf{p}\) for every \(\mathbf{p}\in P_2(\mathbb{R})\), proving spanning.
Independence. Suppose \(u\mathbf{q}_1+v\mathbf{q}_2+w\mathbf{q}_3=0\), the zero polynomial. Its coefficients satisfy \(u+w=0\), \(u+v=0\), and \(v+w=0\). The first two equations minus the third give \(2u=0\), so \(u=0\). Then \(u+v=0\) gives \(v=0\), and \(u+w=0\) gives \(w=0\). Thus the list is independent and is a basis.
Check. For \(\mathbf{p}=2+3x+4x^2\), the formulas give \(u=\frac12\), \(v=\frac52\), and \(w=\frac32\). The combination is \(\frac12(1+x)+\frac52(x+x^2)+\frac32(1+x^2)\). Its constant coefficient is \(2\), its \(x\)-coefficient is \(3\), and its \(x^2\)-coefficient is \(4\), as required.
Proving a Basis for a Subspace
For a subspace, the spanning proof should explicitly use its defining condition. The goal is not to express every vector in the ambient space, but every vector that belongs to the subspace. A good opening is therefore “Let \(\mathbf{x}\) be an arbitrary vector in \(W\),” followed by the equation that membership in \(W\) imposes.
Worked Example: A Candidate Basis for a Subspace of \(\mathbb{R}^4\)
Let \(W=\{(x,y,z,w)\in\mathbb{R}^4:x+y-z=0\}\). Prove that \(\mathbf{b}_1=(-1,1,0,0)\), \(\mathbf{b}_2=(1,0,1,0)\), and \(\mathbf{b}_3=(0,0,0,1)\) form a basis of \(W\).
Spanning. Let \(\mathbf{x}=(x,y,z,w)\in W\). The defining condition gives \(x+y-z=0\), so \(x=-y+z\). Therefore
Every vector in \(W\) has the required representation, so the list spans \(W\).
Independence. Suppose \(a\mathbf{b}_1+b\mathbf{b}_2+c\mathbf{b}_3=\mathbf{0}\). Comparing coordinates yields \(-a+b=0\), \(a=0\), \(b=0\), and \(c=0\). In particular, the second coordinate equation gives \(a=0\); the third gives \(b=0\); and the fourth gives \(c=0\). Thus the only coefficients producing the zero vector are all zero. The list is independent and therefore a basis of \(W\).
Check. The displayed combination has coordinates \((-y+z,y,z,w)\), which satisfy \((-y+z)+y-z=0\), so it belongs to \(W\). The independence calculation also shows that no nonzero choice of coefficients represents \(\mathbf{0}\).
Common Pitfall: Stopping After Spanning
A list that spans a space need not be a basis. For example, consider \((1,0)\), \((0,1)\), and \((1,1)\) in \(\mathbb{R}^2\). The first two vectors already produce every \((r,s)\in\mathbb{R}^2\), since \((r,s)=r(1,0)+s(0,1)\). It would be wrong to conclude that the three-vector list is a basis just because it spans.
The independence check exposes the error. The coefficient choice \(a=1\), \(b=1\), \(c=-1\) gives
This is a nontrivial relation, so the list is dependent and is not a basis. The correction is to state both conclusions separately: “The list spans \(\mathbb{R}^2\), but it is dependent; therefore it is not a basis.” A check on one target vector, or a successful spanning argument, cannot replace the homogeneous coefficient equation required for independence.
A Reliable Proof Checklist
Before finishing a candidate-basis proof, check that each part has the right quantifier and uses the right equation.
- Spanning: Did you take an arbitrary vector in the whole space under consideration, including the subspace condition if there is one?
- Construction: Did you solve for coefficients and verify that they reproduce every coordinate or polynomial coefficient?
- Independence: Did you set a combination equal to the zero vector or zero polynomial and show that every coefficient must vanish?
- Conclusion: Did you explicitly combine the two results to conclude that the list is a basis?
The same coefficient equations often support both parts of a proof, but their roles differ. Solving the equations for an arbitrary target establishes that a representation exists. Solving the homogeneous equations establishes that no nontrivial relation exists. Together these show that every vector in the space has exactly one representation in the candidate list—the defining structure a basis provides.
Check Your Understanding
For each problem, separate the spanning argument from the independence argument, and check the coefficient equations in your proof.
- Prove or disprove that \((1,0)\) and \((0,1)\) form a basis of \(\mathbb{R}^2\).
- For \(\mathbf{u}_1=1+x\), \(\mathbf{u}_2=1-x\), decide whether the list spans \(P_1(\mathbb{R})\) and whether it is independent.
- Let \(W=\{(x,y,z)\in\mathbb{R}^3:x+y+z=0\}\). Test the list \((1,-1,0),(1,0,-1)\) by proving both required claims.
- Explain why a nontrivial solution of \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\) rules out a basis, even if the list spans.
- State what existence and uniqueness of solutions to \(\mathbf{A}\mathbf{c}=\mathbf{b}\) say about the columns of \(\mathbf{A}\).
