Two Requirements for a Basis
A spanning list can describe every vector in a space, but may include redundant vectors. An independent list has no redundancy, but may fail to describe some vectors in the space. A basis is a list that meets both requirements. This definition applies in coordinate spaces and in abstract vector spaces such as spaces of polynomials or matrices.
Throughout, the scalar field is \(\mathbb{R}\). Let \(V\) be a vector space over \(\mathbb{R}\), and let \((\mathbf{v}_1,\dots,\mathbf{v}_p)\) be a finite ordered list in \(V\). Its span, \(\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_p\}\), is the set of all linear combinations \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\), where the coefficients \(c_i\) are real. 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.
The word “list” matters: the vectors have specified positions, and a repeated vector counts as a separate entry. A list with a repeated vector is dependent, since the two copies can be given coefficients \(1\) and \(-1\), with all other coefficients zero. A basis is often described as a set when order is not relevant, but the definition here is applied to an ordered list.
A First Test in \(\mathbb{R}^2\)
Worked Example: A Basis from Two Vectors
Determine whether \(\big((1,1),(1,-1)\big)\) is a basis of \(\mathbb{R}^2\). Write the vectors as columns, as required for vectors in \(\mathbb{R}^2\).
First test whether the list spans. Let \(\mathbf{x}=\begin{bmatrix}x\\y\end{bmatrix}\) be an arbitrary vector in \(\mathbb{R}^2\). We seek scalars \(\alpha,\beta\) satisfying
Comparing coordinates gives \(\alpha+\beta=x\) and \(\alpha-\beta=y\). Adding these equations yields \(2\alpha=x+y\), so \(\alpha=(x+y)/2\). Subtracting the second equation from the first yields \(2\beta=x-y\), so \(\beta=(x-y)/2\). These are real numbers for every real \(x,y\), so the list spans \(\mathbb{R}^2\).
For independence, suppose \(\alpha\begin{bmatrix}1\\1\end{bmatrix}+\beta\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\). The coordinate equations are \(\alpha+\beta=0\) and \(\alpha-\beta=0\). Adding gives \(2\alpha=0\), hence \(\alpha=0\); substituting into the first equation gives \(\beta=0\). The list is independent and spans, so it is a basis.
Check: using the coefficients found for an arbitrary \((x,y)\), the combination has coordinates \(\alpha+\beta=((x+y)+(x-y))/2=x\) and \(\alpha-\beta=((x+y)-(x-y))/2=y\), as required.
This example illustrates the two separate questions in the definition. For spanning, start with an arbitrary vector in the space and ask whether it can be written as a combination of the list. For independence, start with a combination equal to the zero vector and ask whether its coefficients must all vanish. These are different tests; passing one does not establish the other.
Proving What the Definition Implies
Proof. Let \((\mathbf{v}_1,\dots,\mathbf{v}_p)\) be a basis. If \(\mathbf{v}_j=\mathbf{0}\) for some \(j\), assigning coefficient \(1\) to \(\mathbf{v}_j\) and \(0\) to every other vector gives a nontrivial linear combination equal to \(\mathbf{0}\). That contradicts independence. Now suppose that removing \(\mathbf{v}_j\) still left a list spanning \(V\). Since the original list spans \(V\), \(\mathbf{v}_j\) belongs to \(V\), so it could be written as a linear combination of the remaining vectors. Moving that combination to the left would give a relation equal to \(\mathbf{0}\) with coefficient \(1\) on \(\mathbf{v}_j\). This is a nontrivial relation, again contradicting independence. Therefore removing any vector destroys spanning. \(\square\)
The proposition explains why both parts of the definition are needed: independence prevents an entry from being unnecessary in a spanning list. It also gives useful warning signs. A list containing the zero vector cannot be a basis, and a spanning list with a vector that can be removed while preserving its span cannot be a basis.
Worked Example: A Spanning List That Is Not a Basis
Classify the list \(\left(\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix},\begin{bmatrix}1\\1\end{bmatrix}\right)\) in \(\mathbb{R}^2\).
The first two vectors span \(\mathbb{R}^2\): an arbitrary \(\begin{bmatrix}x\\y\end{bmatrix}\) equals \(x\begin{bmatrix}1\\0\end{bmatrix}+y\begin{bmatrix}0\\1\end{bmatrix}\). Therefore the full list spans as well. But the list is dependent, since
The coefficients \(1,1,-1\) are not all zero, so this is a nontrivial dependence relation. The list spans but is not independent; it is not a basis.
Check: the relation gives \(\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}+\begin{bmatrix}0\\1\end{bmatrix}\), so deleting the third vector leaves the span unchanged, in agreement with the proposition.
Bases in Polynomial Spaces
For a polynomial space, spanning means producing every polynomial in that space, not every polynomial of any degree. Write \(P_2(\mathbb{R})\) for the real polynomials of degree at most \(2\), including the zero polynomial. To check spanning, take an arbitrary polynomial \(c+bx+ax^2\) in this space and compare its coefficients with a proposed linear combination. To check independence, set a combination equal to the zero polynomial and use the fact that a polynomial is zero exactly when all its coefficients are zero.
Worked Example: A Basis for \(P_2(\mathbb{R})\)
Determine whether the list \((1+x,\;x+x^2,\;1+x^2)\) is a basis of \(P_2(\mathbb{R})\).
Let \(c+bx+ax^2\) be an arbitrary member of \(P_2(\mathbb{R})\). A combination of the proposed list has the form
Matching coefficients requires \(\alpha+\gamma=c\), \(\alpha+\beta=b\), and \(\beta+\gamma=a\). From the second equation, \(\beta=b-\alpha\). From the third, \(\gamma=a-\beta=a-b+\alpha\). Substitution into the first gives \(2\alpha+a-b=c\), hence
These values exist for every \(a,b,c\in\mathbb{R}\), so the list spans \(P_2(\mathbb{R})\). To check independence, set \(c=b=a=0\) in the same coefficient equations. The formulas then give \(\alpha=\beta=\gamma=0\), so the only combination equal to the zero polynomial is the trivial one. The list is independent and spans; it is a basis.
Check: substituting the formulas into the constant coefficient gives \(\alpha+\gamma=(b+c-a+a-b+c)/2=c\). The coefficient of \(x\) is \(\alpha+\beta=(b+c-a+a+b-c)/2=b\), and that of \(x^2\) is \(\beta+\gamma=(a+b-c+a-b+c)/2=a\).
The same reasoning applies to other polynomial lists: match coefficients to test whether arbitrary elements can be produced, then examine the zero combination for independence. The degree bound is part of the space being tested. For example, proving that a list spans \(P_2(\mathbb{R})\) means representing every polynomial of degree at most \(2\), not polynomials of higher degree.
Bases in Matrix Spaces
The space \(\mathbb{R}^{2\times2}\) consists of all real \(2\times2\) matrices, with addition and scalar multiplication performed entry by entry. A proposed list spans this space if every matrix \(\begin{bmatrix}r&s\\t&u\end{bmatrix}\) can be expressed as a linear combination of the list. For independence, set the combination equal to the zero matrix and require all four entries to be zero.
Worked Example: A Basis for the \(2\times2\) Matrix Space
Classify the following list in \(\mathbb{R}^{2\times2}\):
Take an arbitrary matrix \(\mathbf{X}=\begin{bmatrix}r&s\\t&u\end{bmatrix}\). For scalars \(\alpha,\beta,\gamma,\delta\), entrywise addition gives
To make this equal to \(\mathbf{X}\), the lower-right entry subtracted from the upper-left entry gives \(\delta=r-u\). Then the upper-right and lower-left equations give \(\beta=s-\delta=s-r+u\) and \(\gamma=t-\delta=t-r+u\). Finally, the lower-right equation gives \(\alpha=u-\beta-\gamma=2r-s-t-u\). Thus every \(\mathbf{X}\) is a combination of the four matrices, and the list spans.
These same equations establish independence: when \(\mathbf{X}=\mathbf{O}\), we have \(r=s=t=u=0\), so \(\delta=0\), \(\beta=0\), \(\gamma=0\), and \(\alpha=0\). Therefore the only combination equal to \(\mathbf{O}\) is the trivial one. The list is independent and spans \(\mathbb{R}^{2\times2}\), so it is a basis.
Check: substituting the coefficients into the four entries gives \(\alpha+\beta+\gamma+\delta=r\), \(\beta+\delta=s\), \(\gamma+\delta=t\), and \(\alpha+\beta+\gamma=u\). Hence the resulting matrix is exactly \(\mathbf{X}\).
Common Mistakes When Classifying a List
Wrong calculation: “The three vectors \(\begin{bmatrix}1\\0\end{bmatrix}\), \(\begin{bmatrix}0\\1\end{bmatrix}\), and \(\begin{bmatrix}1\\1\end{bmatrix}\) span \(\mathbb{R}^2\), so they are a basis.” This concludes too much from spanning alone. The correction is to test independence separately: the nontrivial relation displayed earlier shows that the list is dependent, so it is not a basis.
The reverse error is also possible. The one-element list \(\left(\begin{bmatrix}1\\0\end{bmatrix}\right)\) in \(\mathbb{R}^2\) is independent: \(c\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\) forces \(c=0\). But it does not span \(\mathbb{R}^2\), because every combination has second coordinate \(0\), whereas \(\begin{bmatrix}0\\1\end{bmatrix}\) has second coordinate \(1\). It is not a basis of \(\mathbb{R}^2\).
For a candidate list, keep the logical claims separate:
- To prove spanning, represent an arbitrary vector or object in the stated space.
- To prove independence, begin with a combination equal to the zero vector, zero polynomial, or zero matrix and show every coefficient is zero.
- To disprove either condition, give a specific vector that cannot be represented, or a specific nontrivial dependence relation.
There is one small boundary case. The empty list has span \(\{\mathbf{0}\}\), because its only linear combination is the empty sum, defined to be \(\mathbf{0}\); it is also independent, since it has no coefficients that could violate independence. Thus it is a basis of the zero vector space \(\{\mathbf{0}\}\). It cannot span a space containing any nonzero vector.
Check Your Understanding
For each candidate, decide whether it is a basis of the stated space. In every case, account for both spanning and independence.
- Is \(\left(\begin{bmatrix}1\\2\end{bmatrix},\begin{bmatrix}2\\1\end{bmatrix}\right)\) a basis of \(\mathbb{R}^2\)?
- Does \((1,\;1+x,\;x^2)\) form a basis of \(P_2(\mathbb{R})\)?
- Give a nontrivial dependence relation for the list \((1+x,\;2+2x)\) in \(P_2(\mathbb{R})\).
- Can a list containing the zero matrix be a basis of \(\mathbb{R}^{2\times2}\)? Explain using the definition.
- Why does an independent list in a vector space fail to be a basis if it does not span that space?
