Two Obligations for a Basis
A candidate list may look suitable for describing a vector space but still fail in one of two ways. It may contain no redundant vectors and yet miss some vectors in the space. Or it may produce every vector while including redundancy. To establish that a list is a basis, both issues must be addressed: prove that the list is linearly independent, and prove that it spans the space.
Throughout, spaces and scalars are over \(\mathbb{R}\). A linear combination of a finite ordered list \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_p)\) is an expression \(c_1\mathbf{b}_1+\cdots+c_p\mathbf{b}_p\), where each \(c_i\in\mathbb{R}\). Its span is the set of all such linear combinations. The list is linearly independent if the equation \(c_1\mathbf{b}_1+\cdots+c_p\mathbf{b}_p=\mathbf{0}\) forces every coefficient to be zero. These are properties of the list: repeating an entry, for example, can create dependence even though a set would not record that repetition.
Each condition has a different proof goal. For independence, begin with a general zero combination and show that all its coefficients must vanish. For spanning, take an arbitrary vector of the space and produce coefficients that express it using the list. Checking only a few vectors can illustrate the idea, but it does not prove spanning: that claim concerns every vector in \(V\).
A Complete Basis Verification in \(\mathbb{R}^2\)
Worked Example: A Basis with Explicit Coordinates
Show that \(\mathcal{B}=(\mathbf{b}_1,\mathbf{b}_2)\), where \(\mathbf{b}_1=(1,1)\) and \(\mathbf{b}_2=(1,-1)\), is a basis of \(\mathbb{R}^2\). Then find the coordinates of \(\mathbf{x}=(3,1)\) relative to this basis.
Independence. Suppose \(a\mathbf{b}_1+b\mathbf{b}_2=\mathbf{0}\). Comparing the two coordinates gives
Adding the equations gives \(2a=0\), so \(a=0\). Substituting into \(a+b=0\) gives \(b=0\). Thus the zero combination has only zero coefficients, and the list is linearly independent.
Spanning. Let \((x,y)\) be any vector in \(\mathbb{R}^2\). To write it as \(a\mathbf{b}_1+b\mathbf{b}_2\), we need \(a+b=x\) and \(a-b=y\). Adding these equations gives \(2a=x+y\), and subtracting the second from the first gives \(2b=x-y\). Therefore choose
These choices work because \(a+b=(x+y+x-y)/2=x\) and \(a-b=(x+y-x+y)/2=y\). Since \((x,y)\) was arbitrary, the list spans \(\mathbb{R}^2\). Together with independence, this proves it is a basis.
For \(\mathbf{x}=(3,1)\), the formulas give \(a=(3+1)/2=2\) and \(b=(3-1)/2=1\). Hence \([\mathbf{x}]_{\mathcal{B}}=(2,1)^{\top}\). Check: \(2(1,1)+1(1,-1)=(2+1,2-1)=(3,1)\), as required.
The structure of this verification is reusable: the independence argument starts with arbitrary coefficients, while the spanning argument starts with an arbitrary target vector. In the spanning calculation, finding coefficients is not enough on its own; substitute them into the combination and verify that the result is the original arbitrary vector.
What Coordinates Say About Independence
Proof. Suppose first that \(\mathcal{B}\) is a basis. Since it spans \(V\), every \(\mathbf{v}\in V\) has at least one such representation. To show uniqueness, suppose \(\mathbf{v}=\sum_{i=1}^p c_i\mathbf{b}_i=\sum_{i=1}^p d_i\mathbf{b}_i\). Subtracting the two expressions gives \(\sum_{i=1}^p(c_i-d_i)\mathbf{b}_i=\mathbf{0}\). Independence forces \(c_i-d_i=0\) for each \(i\), so \(c_i=d_i\) for each \(i\). Thus the representation is unique. Conversely, suppose every vector in \(V\) has exactly one representation. Existence of a representation for every vector means the list spans \(V\). For independence, suppose \(\sum_{i=1}^p c_i\mathbf{b}_i=\mathbf{0}\). The zero vector also has the representation \(\sum_{i=1}^p 0\mathbf{b}_i=\mathbf{0}\). By uniqueness, \(c_i=0\) for every \(i\). The list is independent and spans \(V\), so it is a basis. \(\square\)
Once a basis is established, the unique coefficients of a vector are its coordinates relative to that ordered basis. The theorem also explains why both tests matter: spanning gives existence of coordinates, and independence gives uniqueness. A list that merely passes one test cannot be called a basis.
Proof. Assume first that \(\mathcal{L}\) is independent, and suppose \(\mathbf{v}=\sum_{i=1}^p a_i\mathbf{u}_i=\sum_{i=1}^p b_i\mathbf{u}_i\). Subtraction gives \(\sum_{i=1}^p(a_i-b_i)\mathbf{u}_i=\mathbf{0}\); independence implies \(a_i-b_i=0\) for every \(i\). Thus the representation is unique. For the reverse direction, suppose \(\mathcal{L}\) is dependent. There are coefficients \(d_1,\dots,d_p\), not all zero, such that \(\sum_{i=1}^p d_i\mathbf{u}_i=\mathbf{0}\). Because \(\mathbf{v}\) belongs to the span, it has some representation \(\mathbf{v}=\sum_{i=1}^p a_i\mathbf{u}_i\). Adding the zero combination gives \(\mathbf{v}=\sum_{i=1}^p(a_i+d_i)\mathbf{u}_i\). This is a different representation, since at least one \(d_i\ne0\). Hence the representation is not unique. \(\square\)
When Independence or Spanning Is Not Enough
Worked Example: Independent but Not Spanning
Consider the one-vector list \(\mathcal{L}=((1,0))\) in \(\mathbb{R}^2\). It is independent: if \(c(1,0)=(0,0)\), then \((c,0)=(0,0)\), which forces \(c=0\). But the list does not span \(\mathbb{R}^2\). Every vector it produces has the form \(c(1,0)=(c,0)\), whose second coordinate is zero. In particular, \((0,1)\) cannot be represented as \(c(1,0)\): the first coordinate would require \(c=0\), while the second would require \(0=1\). So this list satisfies independence but fails spanning, and is not a basis.
Check: the proposed obstruction is an actual vector of the target space, \((0,1)\in\mathbb{R}^2\), and the coordinate equations show that no coefficient works. This is stronger than observing that the list has only one vector: it exhibits precisely what spanning fails to produce.
Worked Example: Spanning but Dependent
Consider \(\mathcal{L}=(\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3)\) in \(\mathbb{R}^2\), where \(\mathbf{u}_1=(1,0)\), \(\mathbf{u}_2=(0,1)\), and \(\mathbf{u}_3=(1,1)\). For any \((x,y)\in\mathbb{R}^2\), the equality \((x,y)=x\mathbf{u}_1+y\mathbf{u}_2+0\mathbf{u}_3\) proves that the list spans \(\mathbb{R}^2\).
The list is nevertheless dependent, since \(\mathbf{u}_1+\mathbf{u}_2-\mathbf{u}_3=(1,0)+(0,1)-(1,1)=(0,0)\), and the coefficients \(1,1,-1\) are not all zero. In fact, the redundancy makes representations non-unique. For any real \(t\),
Taking \(t=0\) gives coefficients \((x,y,0)\), while taking \(t=1\) gives \((x-1,y-1,1)\). These are different coefficient lists and both represent \((x,y)\). Check: the difference between the two lists is \((-1,-1,1)\), whose combination is \(-\mathbf{u}_1-\mathbf{u}_2+\mathbf{u}_3=\mathbf{0}\). The list spans but is not a basis because it fails independence.
A Basis in a Polynomial Space
Worked Example: The Standard Powers in \(P_2(\mathbb{R})\)
Let \(P_2(\mathbb{R})\) be the real vector space of polynomials of degree at most \(2\), including the zero polynomial. Show that \(\mathcal{B}=(1,x,x^2)\) is a basis and find the coordinates of \(p(x)=2-3x+x^2\).
For independence, suppose \(a\cdot1+b\cdot x+c\cdot x^2\) is the zero polynomial. The zero polynomial has zero coefficient at each power of \(x\), so comparison of coefficients gives \(a=0\), \(b=0\), and \(c=0\). Thus the list is independent.
For spanning, let \(q(x)\) be any polynomial in \(P_2(\mathbb{R})\). By the definition of this space, it has the form \(q(x)=a+bx+cx^2\) for real numbers \(a,b,c\). Therefore \(q=a\cdot1+b\cdot x+c\cdot x^2\), so the list spans \(P_2(\mathbb{R})\). The two conditions establish that it is a basis. For the given polynomial, the coefficients are \(2,-3,1\), so \([p]_{\mathcal{B}}=(2,-3,1)^{\top}\). Check: \(2\cdot1+(-3)\cdot x+1\cdot x^2=2-3x+x^2=p(x)\).
Common Pitfall: A Partial Check Is Not a Basis Proof
A common error is to verify that a list spans and then conclude that it is a basis, without checking independence. For the spanning list \(((1,0),(0,1),(1,1))\), one might correctly calculate \((x,y)=x(1,0)+y(0,1)+0(1,1)\) and incorrectly conclude “therefore it is a basis.” That calculation proves spanning, but it says nothing about independence. The correction is to test a general zero combination: \(a(1,0)+b(0,1)+c(1,1)=(a+c,b+c)=(0,0)\). Choosing \(a=1\), \(b=1\), and \(c=-1\) satisfies both equations and is a nontrivial dependence relation. The list spans but is not a basis.
The reverse shortcut is equally unreliable: independence alone does not show that every vector in the target space can be expressed. For the list \(((1,0))\), the coefficient equation proves independence, but the vector \((0,1)\) has no representation. A complete basis verification therefore has two distinct parts, and each must be matched to the right quantifier: arbitrary coefficients for independence, arbitrary target vectors for spanning.
Check Your Understanding
For each question, distinguish what must be shown for independence from what must be shown for spanning.
- For \(\mathcal{B}=((1,2),(2,1))\) in \(\mathbb{R}^2\), write the two coefficient equations for a zero combination. Do they force both coefficients to be zero?
- Show directly that \(((1,0),(0,1))\) spans \(\mathbb{R}^2\), starting with an arbitrary vector \((x,y)\).
- Give a nontrivial dependence relation for \(((1,0),(0,1),(2,1))\) in \(\mathbb{R}^2\).
- Explain why spanning gives at least one representation of each vector, while independence rules out two different representations.
- For \(\mathcal{B}=(1,x,x^2)\), what is the coordinate vector of \(4+x-2x^2\) in \(P_2(\mathbb{R})\)? Check by substitution.
