From Basis Vectors to Coordinates
A vector can be described in more than one way. For instance, in \(\mathbb{R}^2\), the vector \((5,1)\) can be built from the vectors \((1,1)\) and \((1,-1)\) using coefficients \(3\) and \(2\). A coordinate vector records those coefficients in a specified order. The order matters: reversing the basis vectors also reverses the positions of their coefficients. The key guarantee is that a basis gives every vector exactly one such description.
We work over a scalar field \(\mathbb{F}\), usually \(\mathbb{R}\) in the examples. A vector space \(V\) is finite-dimensional if it has a finite basis. A list \((\mathbf{b}_1,\dots,\mathbf{b}_n)\) is a basis of \(V\) if it is linearly independent and spans \(V\). Spanning means that every vector in \(V\) is a linear combination of the list; linear independence means that the only linear combination equal to \(\mathbf{0}\) has every coefficient equal to zero. A basis is an ordered list, so its positions are part of the information.
The coordinate vector belongs to \(\mathbb{F}^n\), while \(\mathbf{v}\) belongs to \(V\). The entries \(c_i\) answer a specific question: how much of the \(i\)th basis vector is used in the representation of \(\mathbf{v}\)? Coordinates are therefore meaningful only together with the ordered basis used to define them. The same vector may have different coordinate vectors relative to different bases, and the same list of coefficients can describe different vectors when attached to different basis vectors.
Why a Basis Gives Exactly One Answer
Proof. Because \(\mathcal{B}\) spans \(V\), each \(\mathbf{v}\in V\) has at least one expression \(\mathbf{v}=c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n\). This proves existence. To prove uniqueness, suppose there are two expressions, \(\mathbf{v}=c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n\) and \(\mathbf{v}=d_1\mathbf{b}_1+\cdots+d_n\mathbf{b}_n\). Subtract the second equality from the first: $$ (c_1-d_1)\mathbf{b}_1+\cdots+(c_n-d_n)\mathbf{b}_n=\mathbf{0}. $$ The basis vectors are linearly independent, so every coefficient in this relation is zero. Thus \(c_i-d_i=0\), and \(c_i=d_i\), for each \(i=1,\dots,n\). The two coefficient lists, and therefore the two coordinate vectors, are identical. \(\square\)
This proof uses both parts of the basis definition, for different purposes. Spanning ensures that a representation exists; independence ensures that no two representations disagree. If either part is missing, the guarantee can fail. A list that does not span may leave some vectors unrepresentable. A dependent list can allow more than one representation for a vector. A coordinate vector is thus not defined relative to an arbitrary list: the list must be a basis.
Computing and Checking Coordinates
Worked Example: Coordinates in an Ordered Basis of R²
Let \(\mathcal{B}=(\mathbf{b}_1,\mathbf{b}_2)\), where \(\mathbf{b}_1=(1,1)\) and \(\mathbf{b}_2=(1,-1)\). Find the coordinates of \(\mathbf{v}=(5,1)\) relative to \(\mathcal{B}\).
First, this list is a basis of \(\mathbb{R}^2\). For any \((x,y)\in\mathbb{R}^2\), take \(a=(x+y)/2\) and \(b=(x-y)/2\). Then \(a+b=x\) and \(a-b=y\), so \((x,y)=a(1,1)+b(1,-1)\). Thus the list spans. If \(a(1,1)+b(1,-1)=(0,0)\), then \(a+b=0\) and \(a-b=0\); adding and subtracting these equations gives \(a=0\) and \(b=0\). The list is independent as well.
Write \(\mathbf{v}=a\mathbf{b}_1+b\mathbf{b}_2\). Comparing components gives \(a+b=5\) and \(a-b=1\). Adding the equations gives \(2a=6\), so \(a=3\). Substitution into \(a+b=5\) gives \(b=2\). Therefore
Check: Reconstruct the vector using the coefficients in order: \(3(1,1)+2(1,-1)=(3+2,3-2)=(5,1)=\mathbf{v}\). The coefficients reproduce the target vector.
The check is part of the computation, not an optional extra: it tests whether the coefficients were attached to the intended basis vectors and whether the arithmetic is correct. In an abstract vector space, reconstruction means evaluating the same linear combination of the basis elements; the entries need not be ordinary coordinates in \(\mathbb{R}^n\).
Worked Example: Coordinates of a Symmetric Matrix
Let \(V\) be the vector space of symmetric \(2\times2\) real matrices, and define the ordered list \(\mathcal{C}=(\mathbf{C}_1,\mathbf{C}_2,\mathbf{C}_3)\) by
Every matrix in \(V\) has the form \(\begin{bmatrix}r&s\\s&t\end{bmatrix}\), and it equals \(r\mathbf{C}_1+s\mathbf{C}_2+t\mathbf{C}_3\). The coefficients are forced to be \(r,s,t\) by the entries in positions \((1,1),(1,2),(2,2)\). Thus \(\mathcal{C}\) spans \(V\), and a combination equal to the zero matrix must have all three coefficients zero. So \(\mathcal{C}\) is a basis.
For \(\mathbf{M}=\begin{bmatrix}2&3\\3&5\end{bmatrix}\), the entries give the coefficients \(2,3,5\), in that order. Hence
Check: \(2\mathbf{C}_1+3\mathbf{C}_2+5\mathbf{C}_3 =\begin{bmatrix}2&0\\0&0\end{bmatrix} +\begin{bmatrix}0&3\\3&0\end{bmatrix} +\begin{bmatrix}0&0\\0&5\end{bmatrix} =\begin{bmatrix}2&3\\3&5\end{bmatrix}=\mathbf{M}\).
Order Is Part of the Coordinates
The basis is an ordered list, not merely a collection of vectors. If \(\mathcal{B}=(\mathbf{b}_1,\mathbf{b}_2)\), then the first coordinate multiplies \(\mathbf{b}_1\) and the second multiplies \(\mathbf{b}_2\). Reversing the list changes which basis vector each position refers to. This can change the coordinate vector even though the represented vector stays the same.
Worked Example: Reversing the Basis Order
In \(\mathbb{R}^3\), let \(\mathcal{B}=(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3)\) and let \(\mathcal{D}=(\mathbf{e}_3,\mathbf{e}_1,\mathbf{e}_2)\), where \(\mathbf{e}_1=(1,0,0)\), \(\mathbf{e}_2=(0,1,0)\), and \(\mathbf{e}_3=(0,0,1)\). These are both bases: each list consists of the three standard unit vectors, so every \((x,y,z)\) is obtained by the corresponding coefficients, uniquely. Find the coordinates of \(\mathbf{v}=(2,-1,4)\) relative to each list.
For \(\mathcal{B}\), the coefficients of \(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\) are \(2,-1,4\), respectively. Thus \([\mathbf{v}]_{\mathcal{B}}=\begin{bmatrix}2\\-1\\4\end{bmatrix}\). For \(\mathcal{D}\), the first basis vector is \(\mathbf{e}_3\), the second is \(\mathbf{e}_1\), and the third is \(\mathbf{e}_2\). The coefficients in this order are \(4,2,-1\), so \([\mathbf{v}]_{\mathcal{D}}=\begin{bmatrix}4\\2\\-1\end{bmatrix}\).
Check: For \(\mathcal{B}\), \(2\mathbf{e}_1-\mathbf{e}_2+4\mathbf{e}_3=(2,-1,4)\). For \(\mathcal{D}\), \(4\mathbf{e}_3+2\mathbf{e}_1-\mathbf{e}_2=(2,-1,4)\) as well. The vectors are the same, but their coordinate vectors differ because the basis order differs.
Why a Spanning List Is Not Enough
Worked Example: A Dependent List Does Not Give Unique Coordinates
Consider the list \(\mathcal{L}=((1,2),(2,4))\) in \(\mathbb{R}^2\). It is not a basis because its second vector is twice its first. Examine whether \(\mathbf{v}=(3,6)\) has a unique coefficient representation, and whether \(\mathbf{w}=(1,0)\) can be represented.
For \(\mathbf{v}\), the equation \(a(1,2)+b(2,4)=(3,6)\) gives \(a+2b=3\) and \(2a+4b=6\). The second equation is twice the first, so it adds no new restriction. Letting \(b=t\) gives \(a=3-2t\) for any \(t\in\mathbb{R}\). For example, \((a,b)=(3,0)\) and \((a,b)=(1,1)\) both work.
Check: \(3(1,2)+0(2,4)=(3,6)\), while \(1(1,2)+1(2,4)=(3,6)\). These are different coefficient columns, so \(\mathbf{v}\) has no unique coordinate vector relative to this list.
For \(\mathbf{w}\), the equation \(a(1,2)+b(2,4)=(1,0)\) would require \(a+2b=1\) and \(2a+4b=0\). The second left-hand side is twice the first, so the equations would imply \(0=2\), which is impossible. Thus \(\mathbf{w}\) has no representation at all from \(\mathcal{L}\). Since the list is dependent and does not span \(\mathbb{R}^2\), it fails both parts of the basis requirement.
Common Mistake: Keeping the Coefficients but Reversing Their Meaning
Suppose \(\mathcal{B}=((1,1),(1,-1))\) and \(\mathbf{v}=(5,1)\). A common error is to write \([\mathbf{v}]_{\mathcal{B}}=\begin{bmatrix}2\\3\end{bmatrix}\), perhaps by reversing the coefficients found earlier. This column means \(2(1,1)+3(1,-1)\), not \(3(1,1)+2(1,-1)\). The calculation gives \(2(1,1)+3(1,-1)=(5,-1)\), which is not \((5,1)\). The correction is to match each coefficient to the basis vector in the same position: \([\mathbf{v}]_{\mathcal{B}}=\begin{bmatrix}3\\2\end{bmatrix}\), and \(3(1,1)+2(1,-1)=(5,1)\).
When finding coordinates, keep three pieces of information together: the ordered basis, the coefficient column in that order, and the reconstruction equation. This prevents confusing an element of \(V\) with a column in \(\mathbb{F}^n\), and makes the answer directly checkable. In the zero-dimensional case, \(V=\{\mathbf{0}\}\) has the empty list as its basis; its only vector has the empty coordinate vector, in keeping with the same existence-and-uniqueness statement.
Check Your Understanding
For each question, identify the ordered list being used and check any proposed coordinates by reconstructing the vector.
- State the two properties a list must have to be a basis of a vector space.
- Let \(\mathcal{B}=((1,0),(1,1))\) in \(\mathbb{R}^2\). Find the coordinates of \((4,3)\) relative to \(\mathcal{B}\), then verify them.
- Why does spanning guarantee that coordinates exist, while linear independence guarantees that they are unique?
- For \(\mathcal{D}=((0,1),(1,0))\), find the coordinates of \((6,2)\). Explain how the order determines the answer.
- For the list \(((1,0),(2,0))\), explain why \((3,0)\) has more than one coefficient representation and why the list cannot be a basis of \(\mathbb{R}^2\).
