Coordinates as a Map
A vector has many possible descriptions as a linear combination, but once an ordered basis is fixed, its basis coefficients are unique. This makes it possible to treat coordinates as a function: it takes a vector as input and returns a column of scalars. The key fact in this tutorial is that the output respects the two basic operations in a vector space. Adding vectors adds their coordinate columns, and multiplying a vector by a scalar multiplies its coordinate column by that scalar.
Let \(V\) be a finite-dimensional vector space over a field \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\), and let \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) be an ordered basis of \(V\). A basis is an ordered list that is linearly independent and spans \(V\). Thus every \(\mathbf{v}\in V\) has a unique expression \(\mathbf{v}=c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n\), with \(c_i\in\mathbb{F}\). Its coordinate vector is the column of those coefficients.
A function \(T:V\to W\) between vector spaces over the same field is linear if \(T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})\) and \(T(c\mathbf{v})=cT(\mathbf{v})\) for all \(\mathbf{u},\mathbf{v}\in V\) and all scalars \(c\in\mathbb{F}\). The coordinate map has exactly these properties. The proof depends on uniqueness of basis coefficients, not on a particular choice of coordinates or on a special formula for the basis.
Why Addition and Scaling Work
Proof. Write the unique basis expansions of \(\mathbf{u}\) and \(\mathbf{v}\) as \(\mathbf{u}=\sum_{i=1}^n a_i\mathbf{b}_i\) and \(\mathbf{v}=\sum_{i=1}^n d_i\mathbf{b}_i\), where each \(a_i,d_i\in\mathbb{F}\). Adding these equalities gives \(\mathbf{u}+\mathbf{v}=\sum_{i=1}^n(a_i+d_i)\mathbf{b}_i\). This is a basis expansion of \(\mathbf{u}+\mathbf{v}\); by uniqueness, its \(i\)th coordinate is \(a_i+d_i\). Therefore \([\mathbf{u}+\mathbf{v}]_{\mathcal{B}}=(a_1+d_1,\dots,a_n+d_n)^{\top}=(a_1,\dots,a_n)^{\top}+(d_1,\dots,d_n)^{\top}\), which is the addition identity. For scalar multiplication, write \(\mathbf{v}=\sum_{i=1}^n d_i\mathbf{b}_i\). For any \(c\in\mathbb{F}\), including \(c=0\), multiplication gives \(c\mathbf{v}=\sum_{i=1}^n(cd_i)\mathbf{b}_i\). This is a basis expansion of \(c\mathbf{v}\), so uniqueness gives \([c\mathbf{v}]_{\mathcal{B}}=(cd_1,\dots,cd_n)^{\top}=c(d_1,\dots,d_n)^{\top}=c[\mathbf{v}]_{\mathcal{B}}\). The two identities are precisely the definition of linearity of \(\Phi_{\mathcal{B}}\). \(\square\)
The order of reasoning matters: first combine or scale the vectors in \(V\), then identify the coefficients in the resulting basis expansion. Uniqueness guarantees that these coefficients are the coordinates. The conclusion does not say that a vector and its coordinate column are the same object: the vector belongs to \(V\), while the coordinate column belongs to \(\mathbb{F}^n\).
Proof. For \(p=1\), the identity is the scalar-multiplication identity in the theorem. Suppose it holds for a sum of \(p-1\) scaled vectors. The sum of \(p\) scaled vectors is the sum of \(\sum_{j=1}^{p-1}c_j\mathbf{v}_j\) and \(c_p\mathbf{v}_p\). Applying the addition identity, then the assumed formula for the first \(p-1\) terms and the scalar identity for the last term, gives \(\left[\sum_{j=1}^{p}c_j\mathbf{v}_j\right]_{\mathcal{B}}=\sum_{j=1}^{p-1}c_j[\mathbf{v}_j]_{\mathcal{B}}+c_p[\mathbf{v}_p]_{\mathcal{B}}\). This proves the formula for \(p\), and induction proves it for every positive integer \(p\). \(\square\)
Worked Examples
Worked Example: Addition and Scaling in a Nonstandard Basis
In \(\mathbb{R}^2\), let \(\mathcal{B}=((1,1)^{\top},(1,-1)^{\top})\), \(\mathbf{u}=(3,1)^{\top}\), and \(\mathbf{v}=(1,3)^{\top}\). Find the coordinates of \(\mathbf{u}+\mathbf{v}\) and \(-2\mathbf{u}\) relative to \(\mathcal{B}\).
First, the two basis vectors are linearly independent: if \(a(1,1)^{\top}+b(1,-1)^{\top}=\mathbf{0}\), then \(a+b=0\) and \(a-b=0\), so \(a=b=0\). They span \(\mathbb{R}^2\), since for any \((x,y)^{\top}\), choosing \(a=(x+y)/2\) and \(b=(x-y)/2\) gives \(a(1,1)^{\top}+b(1,-1)^{\top}=(x,y)^{\top}\). Thus \(\mathcal{B}\) is a basis.
For \(\mathbf{u}\), the equations \(a+b=3\) and \(a-b=1\) give \(2a=4\), so \(a=2\), and then \(b=1\). For \(\mathbf{v}\), the equations \(a+b=1\) and \(a-b=3\) give \(2a=4\), so \(a=2\), and then \(b=-1\). Hence \([\mathbf{u}]_{\mathcal{B}}=(2,1)^{\top}\) and \([\mathbf{v}]_{\mathcal{B}}=(2,-1)^{\top}\). The addition identity gives
Indeed, \(\mathbf{u}+\mathbf{v}=(4,4)^{\top}=4(1,1)^{\top}+0(1,-1)^{\top}\), so \([\mathbf{u}+\mathbf{v}]_{\mathcal{B}}=(4,0)^{\top}\), as predicted. For scaling, \(-2[\mathbf{u}]_{\mathcal{B}}=(-4,-2)^{\top}\). The reconstructed vector is \(-4(1,1)^{\top}-2(1,-1)^{\top}=(-6,-2)^{\top}\), which equals \(-2\mathbf{u}\). Both coordinate identities check directly.
Worked Example: Adding Polynomials in a Polynomial Basis
In the real vector space \(P_2\) of polynomials of degree at most two, use the ordered basis \(\mathcal{B}=(1+x,x,x^2)\). Find the coordinates of \(p+q\), where \(p=2+3x+x^2\) and \(q=1-x+2x^2\).
To find the coordinates, write a general basis combination: \(a(1+x)+bx+cx^2=a+(a+b)x+cx^2\). For \(p\), matching coefficients gives \(a=2\), \(a+b=3\), and \(c=1\). Thus \(b=1\) and \([p]_{\mathcal{B}}=(2,1,1)^{\top}\). For \(q\), matching gives \(a=1\), \(a+b=-1\), and \(c=2\). Thus \(b=-2\) and \([q]_{\mathcal{B}}=(1,-2,2)^{\top}\). The coordinate sum is
Checking in the polynomial space, \(p+q=3+2x+3x^2\), while \(3(1+x)-x+3x^2=3+2x+3x^2\). Therefore \([p+q]_{\mathcal{B}}=(3,-1,3)^{\top}\), in agreement with the coordinate identity.
Worked Example: A Linear Combination in \(\mathbb{R}^3\)
Let \(\mathcal{B}=(\mathbf{b}_1,\mathbf{b}_2,\mathbf{b}_3)\) in \(\mathbb{R}^3\), where \(\mathbf{b}_1=(1,0,1)^{\top}\), \(\mathbf{b}_2=(0,1,1)^{\top}\), and \(\mathbf{b}_3=(1,1,0)^{\top}\). For \(\mathbf{u}=(1,2,3)^{\top}\) and \(\mathbf{v}=(0,5,3)^{\top}\), find \([\mathbf{u}+\mathbf{v}]_{\mathcal{B}}\), then check the result.
For a vector \((x,y,z)^{\top}=a\mathbf{b}_1+b\mathbf{b}_2+c\mathbf{b}_3\), the coordinate equations are \(a+c=x\), \(b+c=y\), and \(a+b=z\). Adding the first two equations and subtracting the third gives \(2c=x+y-z\); then \(a=x-c\) and \(b=y-c\). These formulas give a unique solution for every \((x,y,z)^{\top}\), so the list is a basis. For \(\mathbf{u}\), \(c=(1+2-3)/2=0\), \(a=1\), and \(b=2\). For \(\mathbf{v}\), \(c=(0+5-3)/2=1\), \(a=-1\), and \(b=4\). Consequently
Directly, \(\mathbf{u}+\mathbf{v}=(1,7,6)^{\top}\). Reconstructing from the proposed coordinates gives \(0\mathbf{b}_1+6\mathbf{b}_2+\mathbf{b}_3=(0,6,6)^{\top}+(1,1,0)^{\top}=(1,7,6)^{\top}\). Thus the coordinate column is correct.
Common Mistake: Treating Coordinates as Standard Components
Coordinates depend on the ordered basis. In the first example, \([\mathbf{u}]_{\mathcal{B}}=(2,1)^{\top}\) does not mean that \(\mathbf{u}=(2,1)^{\top}\) in standard coordinates. It means \(\mathbf{u}=2(1,1)^{\top}+1(1,-1)^{\top}=(3,1)^{\top}\).
A concrete error is to multiply \(\mathbf{u}=(3,1)^{\top}\) by \(3\), obtain \(3\mathbf{u}=(9,3)^{\top}\), and then report \((9,3)^{\top}\) as its coordinates in \(\mathcal{B}\). That column reconstructs \(9(1,1)^{\top}+3(1,-1)^{\top}=(12,6)^{\top}\), not \((9,3)^{\top}\). The correction is to scale the existing coordinate column: \([3\mathbf{u}]_{\mathcal{B}}=3(2,1)^{\top}=(6,3)^{\top}\). Checking, \(6(1,1)^{\top}+3(1,-1)^{\top}=(9,3)^{\top}\).
Another source of error is changing the order of a basis without changing the coordinate entries. If the basis vectors are swapped, the coefficients must be swapped as well: the coordinate list records which coefficient belongs to which basis vector. The linearity identities always use the same fixed ordered basis for every vector in the equation.
What the Result Lets You Do
Once a basis is fixed, the theorem turns calculations with vectors into componentwise calculations with columns. The corollary shows this for any finite linear combination: apply the coordinate map after combining vectors, or take the same linear combination of their coordinate columns, and the result agrees. This is useful when vectors are complicated objects, such as polynomials: their basis coefficients can be added or scaled without repeatedly rebuilding the vector from scratch.
The statement has a fixed scope. The same ordered basis must be used for all coordinates in an identity, and the scalars must belong to the field over which \(V\) is a vector space. No claim that coordinates are independent of the basis is involved. Rather, for each fixed basis, unique expansions ensure that the coordinate map preserves addition and scalar multiplication.
Check Your Understanding
Use one fixed ordered basis throughout each calculation, and check coordinate answers by reconstructing the vector.
- For \(\mathcal{B}=((1,1)^{\top},(1,-1)^{\top})\), find \([2(3,1)^{\top}-(1,3)^{\top}]_{\mathcal{B}}\) using the coordinates of the two vectors.
- In \(P_2\) with basis \((1+x,x,x^2)\), find \([2p-q]_{\mathcal{B}}\) for \(p=2+3x+x^2\) and \(q=1-x+2x^2\).
- Explain where uniqueness of basis expansions is used in the proof of the addition identity.
- If \([\mathbf{v}]_{\mathcal{B}}=(a,b)^{\top}\), state \([-\mathbf{v}]_{\mathcal{B}}\) and justify it from the scalar identity.
- Give one reason why a coordinate column cannot usually be read as the vector's standard components.
