Nonstandard Coordinates in R2 and R3

Coordinates Depend on the Chosen Basis

A vector in \(\mathbb{R}^2\) or \(\mathbb{R}^3\) is often written using its standard coordinates, but those entries are not the only way to describe it. Choose a different ordered basis, and the same vector is described by a different list of coefficients. Finding those coefficients is a concrete problem: express the vector as a linear combination of the basis vectors, or, equivalently, solve a linear system.

Throughout, scalars are real. An ordered basis \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) of \(\mathbb{R}^n\) is a list that is linearly independent and spans \(\mathbb{R}^n\). To say that it spans means that every vector in \(\mathbb{R}^n\) is a linear combination of the vectors in the list. The order matters: the first coordinate is the coefficient of \(\mathbf{b}_1\), the second is the coefficient of \(\mathbf{b}_2\), and so on.

The entries of \([\mathbf{x}]_{\mathcal{B}}\) are generally not the entries of \(\mathbf{x}\) in standard coordinates. For example, if the first basis vector is \((1,1)\), then multiplying it by \(c_1\) contributes \(c_1\) to both standard-coordinate entries. The coefficients have to be determined from the whole linear combination.

Proof. Because \(\mathcal{B}\) spans \(\mathbb{R}^n\), there are scalars \(c_1,\dots,c_n\) such that \(\mathbf{x}=c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n\), so a coordinate vector exists. To show uniqueness, suppose also that \(\mathbf{x}=d_1\mathbf{b}_1+\cdots+d_n\mathbf{b}_n\). Subtracting these equations gives \((c_1-d_1)\mathbf{b}_1+\cdots+(c_n-d_n)\mathbf{b}_n=\mathbf{0}\). Since the basis vectors are linearly independent, every coefficient \(c_i-d_i\) is zero. Thus \(c_i=d_i\) for every \(i\), and the coordinate vector is unique. \(\square\)

Set Up the Coefficient System

Write each basis vector as a column in standard coordinates. If \(\mathbf{x}=(x_1,\dots,x_n)^\top\), then the equation defining its coordinates is

Each row of this system says that the corresponding standard-coordinate entry of the weighted sum must match the entry of \(\mathbf{x}\). In dimensions two and three, it is usually efficient to solve by row-reducing the augmented matrix \(\begin{bmatrix}\mathbf{b}_1&\cdots&\mathbf{b}_n&\mathbf{x}\end{bmatrix}\). The entries in the last column then give the coefficients when the left side has been reduced to the identity matrix.

Proof. By the definition of matrix multiplication, the equation \(\begin{bmatrix}\mathbf{b}_1&\cdots&\mathbf{b}_n\end{bmatrix}\mathbf{c}=\mathbf{x}\) is exactly the equation \(\mathbf{x}=c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n\). Its solution \(\mathbf{c}\) is therefore the coordinate vector. Each elementary row operation is reversible: swapping rows reverses itself, scaling a row by a nonzero scalar is reversed by scaling by its reciprocal, and adding a multiple of one row to another is reversed by subtracting that multiple. Consequently these operations preserve the set of solutions to the augmented system. The basis theorem just proved gives exactly one solution, so after row reduction the last column records that unique solution. \(\square\)

Coordinates in a Nonorthogonal Basis of \(\mathbb{R}^2\)

This example illustrates why the coefficients must be found together. The first standard-coordinate entry requires \(c_1+c_2=5\), while the second requires \(c_1+2c_2=8\). Neither coefficient can be read off from just one entry of \(\mathbf{x}\).

Coordinates in a Nonorthogonal Basis of \(\mathbb{R}^3\)

The row-reduction steps solve the three equations together. For this example they are \(c_1+c_2=5\), \(c_2+c_3=4\), and \(c_1+c_3=5\). Each equation involves two coefficients, so trying to assign coefficients one at a time without using the full system risks missing a constraint.

Reconstruct a Vector from Its Coordinates

The coordinate equation works in both directions. When a vector is given, solve for its coefficients. When the coefficients are given, form the indicated linear combination. Reconstructing is also a direct way to check a coordinate calculation: the result must be the original vector in standard coordinates.

A Common Mistake: Treating the Basis as Orthogonal

For an orthogonal basis, dot products can be used to isolate individual coefficients. A nonorthogonal basis does not have that property: taking the dot product of \(\mathbf{x}=c_1\mathbf{b}_1+c_2\mathbf{b}_2\) with \(\mathbf{b}_1\) produces \(c_1(\mathbf{b}_1\cdot\mathbf{b}_1)+c_2(\mathbf{b}_2\cdot\mathbf{b}_1)\). The second term remains when the basis vectors are not orthogonal, so dividing by \(\mathbf{b}_1\cdot\mathbf{b}_1\) alone does not generally give \(c_1\).

For the first worked example, a tempting but wrong calculation is to use the separate ratios \(\mathbf{x}\cdot\mathbf{b}_1/(\mathbf{b}_1\cdot\mathbf{b}_1)=13/2\) and \(\mathbf{x}\cdot\mathbf{b}_2/(\mathbf{b}_2\cdot\mathbf{b}_2)=21/5\) as coordinates. These would reconstruct \((107/10,149/10)^\top\), not \((5,8)^\top\). The correction is to solve both coefficient equations together; row reduction gives \((2,3)^\top\), and \(2(1,1)^\top+3(1,2)^\top=(5,8)^\top\).

Keep the roles of the two columns distinct: the vector \(\mathbf{x}\) is written in standard coordinates, while \([\mathbf{x}]_{\mathcal{B}}\) records coefficients relative to the chosen ordered basis. A reliable routine is to write the basis vectors as columns, append \(\mathbf{x}\), row-reduce, and substitute the resulting coefficients back into the linear combination.