One Operator, Different Matrices

One Map, More Than One Matrix

A linear operator is a function from a vector space to itself. Its action on vectors does not depend on how we choose to describe those vectors. A matrix, however, records that action using coordinates, and coordinates depend on the basis. Consequently, the same operator can be represented by different matrices without changing the operator itself.

We will make this distinction concrete for an operator on \(\mathbb{R}^2\). First we will use the standard basis, then two other ordered bases. In each case, the matrix will be built from the coordinates of the images of the basis vectors. The order of the basis matters: it determines the order of the coordinates and hence the order of the matrix columns.

The uniqueness of coordinates follows directly from independence: if \(\mathbf{v}\) had two expansions, subtracting them would give a linear combination of the basis vectors equal to \(\mathbf{0}\); independence forces every coefficient in that difference to be zero. In particular, coordinates in a basis are not usually the same as the standard coordinates of a vector. The entries in column \(j\) of \([T]_{\mathcal{B}}\) describe \(T(\mathbf{b}_j)\) using the basis \(\mathcal{B}\), not using the standard basis.

Building the Matrix Column by Column

Consider the linear operator \(T:\mathbb{R}^2\to\mathbb{R}^2\) defined by \(T(x,y)=(2x+y,\;x+2y)\). It is linear because each output coordinate is a linear combination of \(x\) and \(y\). The operator \(T\) is the rule itself; its matrix is a coordinate description of that rule in a specified basis.

Now keep the operator fixed and change the ordered basis. To find the new matrix, we still apply \(T\) to each basis vector. The difference is that we must then express each image as a combination of the new basis vectors.

The two matrices found so far are different, but both describe \(T\). They differ because their entries answer different coordinate questions. The standard-basis matrix uses \(\mathcal{E}\)-coordinates for both the basis vectors being acted on and their images. The \(\mathcal{B}\)-matrix uses \(\mathcal{B}\)-coordinates instead.

A Second Coordinate Description

A further basis gives another representation of the same operator. This provides a useful check that the method is about the chosen ordered basis, not about a special property of the first nonstandard basis.

For any basis \(\mathcal{B}\), matrix multiplication has the defining coordinate property \([T(\mathbf{v})]_{\mathcal{B}}=[T]_{\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}\). The next result proves why the column construction gives this property for every vector, rather than only for the basis vectors.

Proof. Write \([\mathbf{v}]_{\mathcal{B}}=(c_1,\dots,c_n)^{\top}\). By the definition of coordinates, \(\mathbf{v}=\sum_{j=1}^{n}c_j\mathbf{b}_j\). Linearity gives \(T(\mathbf{v})=\sum_{j=1}^{n}c_jT(\mathbf{b}_j)\). Taking \(\mathcal{B}\)-coordinates, which respects sums and scalar multiples, gives \([T(\mathbf{v})]_{\mathcal{B}}=\sum_{j=1}^{n}c_j[T(\mathbf{b}_j)]_{\mathcal{B}}\). The right-hand side is exactly the matrix with these coordinate columns multiplied by \((c_1,\dots,c_n)^{\top}\), proving the asserted property. For uniqueness, suppose a matrix \(\mathbf{M}\) also has the property for every \(\mathbf{v}\). Set \(\mathbf{v}=\mathbf{b}_j\). Its coordinate vector is the \(j\)th standard coordinate vector, so \(\mathbf{M}[\mathbf{b}_j]_{\mathcal{B}}\) is column \(j\) of \(\mathbf{M}\). The property says this column equals \([T(\mathbf{b}_j)]_{\mathcal{B}}\). This holds for every \(j\), so \(\mathbf{M}\) has exactly the constructed columns and must equal \([T]_{\mathcal{B}}\). \(\square\)

Checking the Same Input in Different Coordinates

The representation formula can also be checked on a vector that is not a basis vector. This is particularly useful for catching a mismatch between a coordinate vector and the actual vector it describes.

Proof. If \(T=S\), then \(T(\mathbf{b}_j)=S(\mathbf{b}_j)\) for every basis vector, so the coordinate columns defining their matrices agree and \([T]_{\mathcal{B}}=[S]_{\mathcal{B}}\). Conversely, suppose \([T]_{\mathcal{B}}=[S]_{\mathcal{B}}\). For any \(\mathbf{v}\in V\), the theorem gives \([T(\mathbf{v})]_{\mathcal{B}}=[T]_{\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}=[S]_{\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}=[S(\mathbf{v})]_{\mathcal{B}}\). Uniqueness of coordinates implies \(T(\mathbf{v})=S(\mathbf{v})\). This holds for every \(\mathbf{v}\), so \(T=S\). \(\square\)

Common Pitfall: Mixing Coordinate Systems

A frequent error is to compute \(T(\mathbf{b}_j)\) in standard coordinates and put those entries directly into a matrix claimed to be \([T]_{\mathcal{B}}\). For the basis \(\mathcal{B}=((1,0),(1,1))\), this would incorrectly use \((2,1)^{\top}\) as the first column because \(T(1,0)=(2,1)\) in standard coordinates. But that column would assert that \(T(\mathbf{b}_1)=2\mathbf{b}_1+1\mathbf{b}_2=(3,1)\), which is false. The correct \(\mathcal{B}\)-coordinates are \((1,1)^{\top}\), since \(1\mathbf{b}_1+1\mathbf{b}_2=(2,1)\).

The check is straightforward: multiply a proposed matrix by the coordinate vector of a basis vector, then reconstruct the resulting vector using that basis. If the reconstructed vector is not the actual image under \(T\), the column is wrong. This keeps three objects distinct: the operator \(T\), a vector such as \(T(\mathbf{b}_j)\), and the coordinate column representing that vector in a chosen basis.