One Vector, Different Coordinate Columns
A vector does not have just one coordinate column. Its coordinates depend on the ordered basis used to describe it. For example, the vector \((5,1)\) has standard coordinates \(\begin{bmatrix}5\\1\end{bmatrix}\), but its coordinates relative to another basis may be a different column. Both columns can represent the same vector, provided each is interpreted using its own basis.
This distinction is especially useful when a linear map \(T:V\to W\) is being studied with one basis for the domain \(V\) and a different basis for the codomain \(W\). The input vector is reconstructed from its domain coordinates; the output vector is reconstructed from its codomain coordinates. The two coordinate columns are descriptions of vectors, not the vectors themselves. Throughout, examples use real finite-dimensional spaces, and vectors and coordinate columns are written as columns.
Coordinates Relative to an Ordered Basis
Let \(V\) be a vector space over \(\mathbb{R}\). A basis \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) is an ordered list that spans \(V\) and is linearly independent. Spanning means that every vector in \(V\) is a linear combination of the listed vectors; linear independence means that the only linear combination equal to \(\mathbf{0}\) has all coefficients zero. The order matters because the first coefficient is recorded in the first position, the second in the second, and so on.
The existence and uniqueness in this definition follow directly from the two parts of the basis condition. Spanning supplies at least one list of coefficients. If two lists of coefficients both represent \(\mathbf{v}\), subtracting the representations gives a linear combination of the basis vectors equal to \(\mathbf{0}\); independence forces the two lists to agree.
Proof. Since \(\mathcal{B}\) spans \(V\), there are scalars \(c_1,\dots,c_n\) such that \(\mathbf{v}=c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n\). This gives existence. To prove uniqueness, suppose also that \(\mathbf{v}=d_1\mathbf{b}_1+\cdots+d_n\mathbf{b}_n\). Subtracting the two equations gives \((c_1-d_1)\mathbf{b}_1+\cdots+(c_n-d_n)\mathbf{b}_n=\mathbf{0}\). Since the basis is linearly independent, \(c_i-d_i=0\) for every \(i\). Thus \(c_i=d_i\) for every \(i\), so the coordinate column is unique. \(\square\)
For the standard basis of \(\mathbb{R}^n\), the coordinate column is just the familiar column of components. With a different basis, those familiar components usually do not give the coordinates. To find coordinates, express the vector as a linear combination of the specified basis vectors and use the coefficients, in order, as the entries of the coordinate column.
Computing Coordinates in Two Bases
Worked Example: Domain and Codomain Bases in \(\mathbb{R}^2\)
Let \(\mathcal{B}=((1,1),(1,-1))\) be a basis for the domain \(\mathbb{R}^2\), and let \(\mathcal{C}=((1,0),(1,2))\) be a basis for the codomain \(\mathbb{R}^2\). Consider \(T(x_1,x_2)=(x_1+2x_2,3x_1-x_2)\) and the input \(\mathbf{x}=(5,1)\). Find the input coordinates in \(\mathcal{B}\) and the output coordinates in \(\mathcal{C}\).
First, both lists are bases. For \(\mathcal{B}\), a vector \((r,s)\) has a representation \(a(1,1)+b(1,-1)\) when \(a+b=r\) and \(a-b=s\). These equations have the unique solution \(a=(r+s)/2\), \(b=(r-s)/2\), so the list spans and has unique coefficients. For \(\mathcal{C}\), \(a(1,0)+b(1,2)=(a+b,2b)\). To represent \((r,s)\), take \(b=s/2\) and \(a=r-s/2\); these are also unique coefficients.
Write the input as a combination of the domain basis vectors. The equations are \(a+b=5\) and \(a-b=1\). Adding them gives \(2a=6\), so \(a=3\). Then \(3+b=5\), so \(b=2\). Therefore
Now compute the actual output vector: \(T(5,1)=(5+2,\;3(5)-1)=(7,14)\). To express it in the codomain basis, solve \(a(1,0)+b(1,2)=(7,14)\). The second component gives \(2b=14\), hence \(b=7\); the first gives \(a+b=7\), hence \(a=0\). Thus
Check: Reconstructing from \([\mathbf{x}]_{\mathcal{B}}\) gives \((5,1)\), and reconstructing from \([T(\mathbf{x})]_{\mathcal{C}}\) gives \((7,14)\), exactly the output computed from the formula for \(T\). The input and output coordinate columns are not the same, even though both are columns with two entries: they refer to different vectors and use different bases.
Coordinates Are Coefficients, Not Components by Default
The standard coordinate column \(\begin{bmatrix}r\\s\end{bmatrix}\) describes \((r,s)\) using the standard basis \(((1,0),(0,1))\). If the basis changes, the coefficients needed to build the vector change too. A reliable procedure is to write an equation using the specified basis vectors, equate components, solve for the coefficients, and substitute those coefficients back into the combination.
Proof. Write \(\mathbf{u}=\sum_{i=1}^{n}u_i\mathbf{b}_i\) and \(\mathbf{v}=\sum_{i=1}^{n}v_i\mathbf{b}_i\), where \(u_i\) and \(v_i\) are their coordinates relative to \(\mathcal{B}\). Then \(a\mathbf{u}+b\mathbf{v}=\sum_{i=1}^{n}(au_i+bv_i)\mathbf{b}_i\). By the definition of coordinates, the coefficient column for this vector is \(\begin{bmatrix}au_1+bv_1\\ \vdots\\ au_n+bv_n\end{bmatrix}\), which equals \(a[\mathbf{u}]_{\mathcal{B}}+b[\mathbf{v}]_{\mathcal{B}}\). Uniqueness of coordinates ensures this is the coordinate column of \(a\mathbf{u}+b\mathbf{v}\). \(\square\)
This result concerns coordinates relative to one fixed basis. If the basis changes, the coefficient columns must be found relative to that new basis. The underlying vector remains unchanged; what changes is the list of coefficients used to reconstruct it.
Worked Example: One Polynomial in Two Bases
Work in the real vector space \(P_2\) of polynomials of degree at most \(2\). Consider the ordered bases \(\mathcal{B}=(1+x,x,x^2)\) and \(\mathcal{C}=(1,1+x,x^2)\), and the polynomial \(p(x)=3+5x+2x^2\). Find the two coordinate columns.
Both lists are bases. For \(\mathcal{B}\), \(a(1+x)+bx+cx^2=a+(a+b)x+cx^2\). Any polynomial \(r_0+r_1x+r_2x^2\) is obtained with \(a=r_0\), \(b=r_1-r_0\), and \(c=r_2\), uniquely. For \(\mathcal{C}\), \(a+b(1+x)+cx^2=(a+b)+bx+cx^2\). The coefficients are uniquely \(b=r_1\), \(a=r_0-r_1\), and \(c=r_2\), so this list also spans \(P_2\) with unique coefficients.
For \(\mathcal{B}\), write \(p(x)=a(1+x)+bx+cx^2\). Matching the constant, \(x\), and \(x^2\) coefficients gives \(a=3\), \(a+b=5\), and \(c=2\). Hence \(b=2\), and
For \(\mathcal{C}\), write \(p(x)=a+b(1+x)+cx^2\). Matching coefficients gives \(a+b=3\), \(b=5\), and \(c=2\). Thus \(a=-2\), and
Check: Expanding both basis combinations gives \(3+5x+2x^2\). The different columns are not conflicting descriptions: the first uses \(1+x,x,x^2\), while the second uses \(1,1+x,x^2\).
Solving for Coordinates in Three Dimensions
Worked Example: Coordinates in a Nonstandard Basis of \(\mathbb{R}^3\)
Let \(\mathcal{D}=((1,1,0),(1,0,1),(0,1,1))\). Find the \(\mathcal{D}\)-coordinates of \(\mathbf{v}=(4,2,6)\).
First confirm that \(\mathcal{D}\) is a basis. For an arbitrary vector \((r,s,t)\), the equation \(a(1,1,0)+b(1,0,1)+c(0,1,1)=(r,s,t)\) gives \(a+b=r\), \(a+c=s\), and \(b+c=t\). Adding the first two equations and subtracting the third gives \(2a=r+s-t\). The corresponding solutions are
These formulas give coefficients for every \((r,s,t)\), and the equations determine them uniquely. Thus the list spans \(\mathbb{R}^3\) and is linearly independent. For \((r,s,t)=(4,2,6)\), they give \(a=(4+2-6)/2=0\), \(b=(4+6-2)/2=4\), and \(c=(2+6-4)/2=2\). Therefore
Check: The first component is \(0+4+0=4\), the second is \(0+0+2=2\), and the third is \(0+4+2=6\). These match \(\mathbf{v}\).
Common Mistake: Copying the Vector’s Components
For \(\mathbf{x}=(5,1)\) in the first example, a tempting but incorrect answer is \([\mathbf{x}]_{\mathcal{B}}=\begin{bmatrix}5\\1\end{bmatrix}\), copied from the standard components of \(\mathbf{x}\). That column means \(5(1,1)+1(1,-1)\) when interpreted in \(\mathcal{B}\), which equals \((6,4)\), not \((5,1)\). The correction is to solve \(\mathbf{x}=a(1,1)+b(1,-1)\), giving \(a+b=5\), \(a-b=1\), and therefore \([\mathbf{x}]_{\mathcal{B}}=\begin{bmatrix}3\\2\end{bmatrix}\). Substitution verifies that \(3(1,1)+2(1,-1)=(5,1)\).
Another useful check is to keep track of the basis alongside every coordinate column. A column such as \(\begin{bmatrix}3\\2\end{bmatrix}\) alone does not identify a vector in an abstract space: it needs the instruction to use \(\mathcal{B}\), or another specified basis, to reconstruct the vector.
Using Separate Bases for Inputs and Outputs
When a map \(T:V\to W\) is under consideration, the input lies in \(V\) and its image lies in \(W\). If \(\mathcal{B}\) is a basis of \(V\) and \(\mathcal{C}\) is a basis of \(W\), then \([\mathbf{v}]_{\mathcal{B}}\) describes \(\mathbf{v}\) using vectors in the domain, while \([T(\mathbf{v})]_{\mathcal{C}}\) describes the image using vectors in the codomain. Even when \(V\) and \(W\) are both \(\mathbb{R}^2\), the chosen bases need not be the same. The worked example showed that the domain coordinates \(\begin{bmatrix}3\\2\end{bmatrix}\) reconstruct \((5,1)\), whereas the codomain coordinates \(\begin{bmatrix}0\\7\end{bmatrix}\) reconstruct \(T(5,1)=(7,14)\).
To compute either column reliably, follow the same sequence: identify the vector and its space, write a linear combination of that space’s specified basis vectors, solve for the coefficients, place them in basis order, and reconstruct the vector as a check. Keeping these steps separate prevents a coordinate column from being mistaken for either the original vector or its image.
Check Your Understanding
For each question, state the basis being used and check a computed coordinate column by reconstructing its vector.
- For \(\mathcal{B}=((1,1),(1,-1))\), find the coordinates of \((2,4)\).
- In \(P_2\), find the coordinates of \(1+4x+3x^2\) relative to \((1+x,x,x^2)\).
- Verify that \((1,0,1),(0,1,1),(1,1,0)\) form a basis of \(\mathbb{R}^3\), then find the coordinates of \((3,5,4)\).
- Explain why two different coordinate columns can represent the same polynomial in two different bases.
- For a map \(T:V\to W\), identify which basis is used to compute \([\mathbf{v}]_{\mathcal{B}}\) and which is used to compute \([T(\mathbf{v})]_{\mathcal{C}}\).
