Why Choose a Basis Around the Vectors You Use?
Coordinates depend on the basis. In the standard basis of \(\mathbb{R}^n\), the coordinate vector of \(\mathbf{x}\) is simply its column of entries. A different basis may make the coordinates of particular vectors much shorter or more meaningful. For example, if an important vector is itself the first basis vector, its coordinates are \((1,0,\dots,0)^{\top}\), even if its standard coordinates are less simple.
Choosing such a basis does not change the vector. It changes the ordered list of vectors used to describe it. The choice is useful when a problem repeatedly involves a particular vector or an independent list: include those vectors as basis vectors, then complete the list so that it still spans the whole space. Throughout this tutorial the scalars are real, and the spaces considered are finite-dimensional subspaces of \(\mathbb{R}^n\).
For comparison, the standard basis is \(\mathcal{E}=(\mathbf{e}_1,\dots,\mathbf{e}_n)\), where each \(\mathbf{e}_i\) has a \(1\) in position \(i\) and \(0\) elsewhere. Thus \([\mathbf{x}]_{\mathcal{E}}=\mathbf{x}\). A basis adapted to a vector \(\mathbf{v}\ne\mathbf{0}\) can place \(\mathbf{v}\) first; then \([\mathbf{v}]_{\mathcal{B}}=(1,0,\dots,0)^{\top}\). This is not a shortcut that avoids checking the basis conditions: the full list must still be independent and span the space.
Including a Target Vector in a Basis
A nonzero vector is independent by itself, so it can begin a basis. More generally, a list can be included in a basis if it is independent. The following construction explains how to complete such a list using the standard basis vectors. It also gives a practical method: keep the vectors of interest first, then add standard basis vectors only when they contribute a direction not already spanned.
Proof. Start with the given independent list. Examine \(\mathbf{e}_1,\dots,\mathbf{e}_n\) in order. Whenever the standard basis vector being examined is not in the span of the current list, append it. Appending such a vector preserves independence: if a linear combination of the enlarged list equals \(\mathbf{0}\) and the new vector has a nonzero coefficient, rearranging would express that vector as a combination of the old list, contrary to its not being in the old span. If its coefficient is zero, the remaining relation has only the old vectors, whose independence forces all remaining coefficients to be zero. At the end, every \(\mathbf{e}_i\) is in the span of the final list: it was either appended, or already belonged to the span when examined, which remains contained in the span as more vectors are added. Since every vector in \(\mathbb{R}^n\) is a linear combination of \(\mathbf{e}_1,\dots,\mathbf{e}_n\), the final list spans \(\mathbb{R}^n\). It is both independent and spanning, hence is a basis. If \(p=0\), the same process starts with the empty list and produces the standard basis. \(\square\)
Being independent is essential. If a target list is dependent, it cannot appear in full as part of a basis: every sublist of a basis is independent, whereas the target list already has a nontrivial dependence relation. In that situation, identify an independent sublist that retains the directions needed, and include that sublist instead. A dependent vector can still have simple coordinates in the resulting basis, but it cannot be a basis entry alongside all the vectors that make it dependent.
Coordinates of the Included Vectors
Proof. The vector \(\mathbf{b}_j\) has the representation \(0\mathbf{b}_1+\cdots+0\mathbf{b}_{j-1}+1\mathbf{b}_j+0\mathbf{b}_{j+1}+\cdots+0\mathbf{b}_n\). By uniqueness of the representation in a basis, its coordinate vector is exactly the column with \(1\) in position \(j\) and zero in every other position, namely \(\mathbf{e}_j\). \(\square\)
This proposition describes one immediate benefit of an adapted basis. The benefit extends to other vectors too: if a vector is a short linear combination of the chosen basis vectors, its coordinates are precisely the coefficients in that combination. The examples below compare those coefficients with the standard coordinates, and check each proposed basis and coordinate calculation directly.
Worked Example: Making One Vector the First Basis Vector
Choose a basis of \(\mathbb{R}^2\) beginning with \(\mathbf{v}=(3,2)^{\top}\). Then find the coordinates of \(\mathbf{x}=(11,8)^{\top}\) in that basis and compare them with standard coordinates.
Choose \(\mathbf{b}_2=(1,1)^{\top}\), so \(\mathcal{B}=(\mathbf{v},\mathbf{b}_2)\). To check independence, suppose \(a\mathbf{v}+b\mathbf{b}_2=\mathbf{0}\). This gives \(3a+b=0\) and \(2a+b=0\). Subtracting the second equation from the first gives \(a=0\); then \(b=0\). Thus the two vectors are independent. In \(\mathbb{R}^2\), two independent vectors span \(\mathbb{R}^2\), so this is a basis.
To find \([\mathbf{x}]_{\mathcal{B}}\), solve \(a\mathbf{v}+b\mathbf{b}_2=\mathbf{x}\):
Subtracting the second equation from the first gives \(a=3\). Substitution into \(2a+b=8\) gives \(6+b=8\), so \(b=2\). Therefore \([\mathbf{x}]_{\mathcal{B}}=(3,2)^{\top}\), whereas \([\mathbf{x}]_{\mathcal{E}}=(11,8)^{\top}\).
Check: \(3(3,2)^{\top}+2(1,1)^{\top}=(9+2,6+2)^{\top}=(11,8)^{\top}\). Also, \([\mathbf{v}]_{\mathcal{B}}=(1,0)^{\top}\), as expected.
Worked Example: Preserving an Important Independent Pair
Extend \(\mathbf{u}=(1,1,0)^{\top}\) and \(\mathbf{v}=(1,0,1)^{\top}\) to a basis of \(\mathbb{R}^3\), then find the coordinates of \(\mathbf{x}=(5,2,1)^{\top}\) in that basis.
The pair is independent: if \(a\mathbf{u}+b\mathbf{v}=\mathbf{0}\), then \((a+b,a,b)^{\top}=\mathbf{0}\). The second and third entries give \(a=0\) and \(b=0\). Examine the standard basis vectors to complete the pair. The vector \(\mathbf{e}_1=(1,0,0)^{\top}\) is not in the span of \(\mathbf{u},\mathbf{v}\): an equation \(a\mathbf{u}+b\mathbf{v}=\mathbf{e}_1\) would require \(a=0\) from the second entry and \(b=0\) from the third, but then the first entry would be \(0\), not \(1\). Append \(\mathbf{e}_1\) to obtain \(\mathcal{B}=(\mathbf{u},\mathbf{v},\mathbf{e}_1)\). Its independence can also be checked directly: \(a\mathbf{u}+b\mathbf{v}+c\mathbf{e}_1=(a+b+c,a,b)^{\top}=\mathbf{0}\) implies \(a=0\), \(b=0\), and then \(c=0\). It has three independent vectors in \(\mathbb{R}^3\), so it is a basis.
Write the target as a combination of the basis vectors. The coordinate equations are
The second entry gives \(a=2\), and the third gives \(b=1\). The first then gives \(2+1+c=5\), so \(c=2\). Thus \([\mathbf{x}]_{\mathcal{B}}=(2,1,2)^{\top}\), while \([\mathbf{x}]_{\mathcal{E}}=(5,2,1)^{\top}\).
Check: \(2\mathbf{u}+\mathbf{v}+2\mathbf{e}_1=(2,2,0)^{\top}+(1,0,1)^{\top}+(2,0,0)^{\top}=(5,2,1)^{\top}\).
When the Target List Is Dependent
A list may contain useful information without being eligible to appear in full as a basis. The right question is whether every vector in the list contributes a new direction. If one vector is a combination of others, keep an independent subset and express the omitted vector using the basis. This both respects the basis conditions and makes the dependence visible in coordinates.
Worked Example: Keeping a Spanning Pair and Recording a Dependence
Consider \(\mathbf{u}=(1,1,0)^{\top}\), \(\mathbf{v}=(0,1,1)^{\top}\), and \(\mathbf{w}=(1,2,1)^{\top}\). Find a basis containing \(\mathbf{u}\) and \(\mathbf{v}\), and compare the coordinates of \(\mathbf{w}\) and \(\mathbf{x}=(3,5,2)^{\top}\) in that basis with their standard coordinates.
First, \(\mathbf{w}=\mathbf{u}+\mathbf{v}\), so the three-vector list is dependent and cannot be included in full in a basis. The first two vectors are independent: \(a\mathbf{u}+b\mathbf{v}=(a,a+b,b)^{\top}=\mathbf{0}\) gives \(a=0\) and \(b=0\). Append \(\mathbf{e}_3=(0,0,1)^{\top}\). The resulting list \(\mathcal{B}=(\mathbf{u},\mathbf{v},\mathbf{e}_3)\) is independent because \(a\mathbf{u}+b\mathbf{v}+c\mathbf{e}_3=(a,a+b,b+c)^{\top}=\mathbf{0}\) first gives \(a=0\), then \(b=0\), then \(c=0\). It is therefore a basis of \(\mathbb{R}^3\).
Since \(\mathbf{w}=\mathbf{u}+\mathbf{v}\), its adapted coordinates are \([\mathbf{w}]_{\mathcal{B}}=(1,1,0)^{\top}\). Its standard coordinates are \((1,2,1)^{\top}\). For \(\mathbf{x}\), solve \((a,a+b,b+c)^{\top}=(3,5,2)^{\top}\). The first entry gives \(a=3\); the second gives \(3+b=5\), so \(b=2\); the third gives \(2+c=2\), so \(c=0\). Hence \([\mathbf{x}]_{\mathcal{B}}=(3,2,0)^{\top}\), compared with \([\mathbf{x}]_{\mathcal{E}}=(3,5,2)^{\top}\).
Check: \(1\mathbf{u}+1\mathbf{v}+0\mathbf{e}_3=(1,2,1)^{\top}=\mathbf{w}\), and \(3\mathbf{u}+2\mathbf{v}+0\mathbf{e}_3=(3,3,0)^{\top}+(0,2,2)^{\top}=(3,5,2)^{\top}=\mathbf{x}\).
A Practical Choice-and-Check Method
When the goal is to simplify coordinates for one vector or a list, the following process keeps the choice purposeful. The steps apply in \(\mathbb{R}^n\); the list to be preserved must be independent if every member is to be a basis vector.
Put the important vector or independent list first, in the order you want. If the list is dependent, decide which independent vectors retain the directions you need.
Test standard basis vectors one at a time. Append a vector only if it is not already in the span of the current list. This is the construction proved above.
Check independence and spanning, or use the completion argument once the list is known to be independent and to have \(n\) vectors in \(\mathbb{R}^n\).
Set the linear combination of the chosen basis vectors equal to the target vector and solve for the coefficients. Those coefficients, in basis order, form the coordinate column.
Multiply each basis vector by its proposed coordinate and add. The result must be the original vector.
The coordinates are not automatically simpler for every vector. An adapted basis is useful when the vectors that matter have short expressions in it, not simply because it is nonstandard. If a later calculation involves many vectors, consider whether the chosen basis makes those vectors easier to describe while still keeping the basis easy to verify.
Common Mistake: Treating a Dependent List as a Basis
A frequent error is to include every vector of interest without checking independence. For instance, one might claim that \(\mathcal{B}=((1,1)^{\top},(2,2)^{\top})\) is a basis of \(\mathbb{R}^2\) because it has two vectors. That claim is wrong: \((2,2)^{\top}=2(1,1)^{\top}\), and \(2(1,1)^{\top}-1(2,2)^{\top}=(0,0)^{\top}\) is a nontrivial dependence relation.
The correction is to replace the redundant vector with one that adds a new direction. For example, \(((1,1)^{\top},(1,0)^{\top})\) is independent: \(a(1,1)^{\top}+b(1,0)^{\top}=(a+b,a)^{\top}=(0,0)^{\top}\) implies \(a=0\) and \(b=0\). This pair is a basis of \(\mathbb{R}^2\). The check is not optional: having the right number of vectors does not by itself make a list a basis.
Check Your Understanding
For each coordinate claim, write the basis combination and reconstruct the vector to check your answer.
- Choose a basis of \(\mathbb{R}^2\) beginning with \((2,1)^{\top}\). Give one possible second vector and verify independence.
- In the basis \(((1,0,1)^{\top},(0,1,1)^{\top},(0,0,1)^{\top})\), find the coordinates of \((3,2,4)^{\top}\).
- Why can \((1,0)^{\top}\) and \((2,0)^{\top}\) not both appear in a basis of \(\mathbb{R}^2\)?
- Extend the independent list \((1,1,0)^{\top},(0,1,0)^{\top}\) to a basis of \(\mathbb{R}^3\) using a standard basis vector.
- A vector has coordinates \((0,0,1)^{\top}\) in an ordered basis \((\mathbf{b}_1,\mathbf{b}_2,\mathbf{b}_3)\). Express it as a combination of the basis vectors.
