Basis Matrices and Vector Reconstruction

From an Ordered Basis to a Matrix

An ordered basis does more than describe which vectors can be used to build a space. The order tells us which coefficient belongs to which basis vector, so it also gives a consistent way to encode every vector by a column of coordinates. A matrix collects the basis vectors in that same order. Multiplying this matrix by a coordinate column rebuilds the vector from its basis ingredients.

Throughout, the vectors are in \(\mathbb{R}^n\) and are written as columns. An ordered basis of \(\mathbb{R}^n\) is an ordered list \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) that is linearly independent and spans \(\mathbb{R}^n\). Because it spans, every \(\mathbf{v}\in\mathbb{R}^n\) can be written as a linear combination of its vectors; because it is linearly independent, that expression is unique. Its coordinate vector \([\mathbf{v}]_{\mathcal{B}}\) is the column of coefficients in that unique expression.

The column order is essential. If \([\mathbf{v}]_{\mathcal{B}}=(c_1,\dots,c_n)^{\top}\), then the first coefficient multiplies the first column of \(\mathbf{B}\), the second coefficient multiplies the second column, and so on. Thus, matrix multiplication is exactly the operation of forming the corresponding linear combination.

Proof. By the definition of the coordinate vector, there are unique scalars \(c_1,\dots,c_n\) such that \(\mathbf{v}=c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n\), and \([\mathbf{v}]_{\mathcal{B}}=(c_1,\dots,c_n)^{\top}\). By the definition of matrix multiplication, multiplying \(\mathbf{B}\) by this column gives \(c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n\). This equals \(\mathbf{v}\), as required. \(\square\)

Reconstructing a Vector

This calculation goes from basis coordinates to standard coordinates. The reverse task starts with the standard-coordinate column \(\mathbf{v}\) and asks for coefficients \(c_1,\dots,c_n\). The reconstruction theorem says exactly what those coefficients must satisfy: \(\mathbf{B}\mathbf{c}=\mathbf{v}\). In other words, finding basis coordinates is solving a linear system whose coefficient matrix is the basis matrix.

Why the Basis Matrix Has an Inverse

For a basis matrix, the system \(\mathbf{B}\mathbf{c}=\mathbf{v}\) has exactly one solution for every \(\mathbf{v}\in\mathbb{R}^n\). Existence follows because the basis spans the space; uniqueness follows because the basis is linearly independent. These two facts imply that the matrix can be inverted, so one matrix operation can recover coordinates from standard coordinates.

Proof. Define \(T:\mathbb{R}^n\to\mathbb{R}^n\) by \(T(\mathbf{c})=\mathbf{B}\mathbf{c}\). The reconstruction theorem and the fact that the basis spans show that every \(\mathbf{v}\in\mathbb{R}^n\) equals \(T(\mathbf{c})\) for some \(\mathbf{c}\), so \(T\) is onto. If \(T(\mathbf{c})=T(\mathbf{d})\), then \(\mathbf{B}(\mathbf{c}-\mathbf{d})=\mathbf{0}\). This says that a linear combination of the basis vectors is zero, so linear independence forces every entry of \(\mathbf{c}-\mathbf{d}\) to be zero. Thus \(\mathbf{c}=\mathbf{d}\), and \(T\) is one-to-one. Hence \(T\) has an inverse map. Since \(T\) is represented by the square matrix \(\mathbf{B}\), this inverse map is represented by a matrix \(\mathbf{C}\) satisfying \(\mathbf{B}\mathbf{C}=\mathbf{I}_n\) and \(\mathbf{C}\mathbf{B}=\mathbf{I}_n\); this is the definition of \(\mathbf{B}^{-1}\). Now \(\mathbf{B}[\mathbf{v}]_{\mathcal{B}}=\mathbf{v}\). Multiplying by \(\mathbf{B}^{-1}\) gives \([\mathbf{v}]_{\mathcal{B}}=\mathbf{B}^{-1}\mathbf{v}\). \(\square\)

The inverse formula is not a different coordinate convention. It is the reconstruction equation solved for its coordinate column. In computation, one need not explicitly form \(\mathbf{B}^{-1}\): solving \(\mathbf{B}\mathbf{c}=\mathbf{v}\) directly can be more efficient. In either approach, the matrix whose columns are the ordered basis vectors is the matrix that acts on the basis-coordinate column.

Order, Shape, and Interpretation

Three features of the basis-matrix rule are worth keeping distinct.

• Columns represent basis vectors. The \(j\)th column is \(\mathbf{b}_j\) in standard coordinates, not the coordinate vector of an arbitrary \(\mathbf{v}\).
• The input is a coordinate column. The entries of \([\mathbf{v}]_{\mathcal{B}}\) are coefficients, not generally the standard entries of \(\mathbf{v}\).
• Order fixes the coefficient positions. Reordering the basis vectors changes the corresponding basis matrix and also changes which coefficient belongs in each position.

For the standard basis \((\mathbf{e}_1,\dots,\mathbf{e}_n)\), the basis matrix is \(\mathbf{I}_n\). Therefore multiplying by the basis matrix changes nothing, and a vector's standard-coordinate column equals its coordinate column in the standard basis. For a nonstandard basis, the entries of \([\mathbf{v}]_{\mathcal{B}}\) may differ from the standard entries of \(\mathbf{v}\), even though both columns encode the same vector in different ways.

Common Mistake: Putting Basis Vectors in Rows

A frequent error is to put the basis vectors in the rows and then multiply that matrix by the coordinate column. With \(\mathbf{b}_1=(1,0)^{\top}\) and \(\mathbf{b}_2=(2,1)^{\top}\), the basis matrix has these vectors as columns:

For coordinates \((3,-1)^{\top}\), the correct reconstruction is \(\mathbf{B}\begin{bmatrix}3\\-1\end{bmatrix}=\begin{bmatrix}1\\-1\end{bmatrix}\). If the vectors are incorrectly placed in rows, the resulting matrix is \(\begin{bmatrix}1&0\\2&1\end{bmatrix}=\mathbf{B}^{\top}\), and the wrong product is

This does not reconstruct the vector represented by the stated coordinates. The correction is to use the basis vectors as columns, so that matrix multiplication forms \(3\mathbf{b}_1-\mathbf{b}_2\). Indeed, \(3(1,0)^{\top}-(2,1)^{\top}=(1,-1)^{\top}\).

Check the Direction by Reconstructing

A reliable way to check coordinate work is to substitute the proposed coordinate column into the reconstruction equation. If it returns the given standard-coordinate vector, the coefficients are in the correct order and produce the right vector. This check is especially useful when a problem gives several bases or when the coordinate column is found by solving equations.