Matrices of Sums and Scalar Multiples of Maps

From Operations on Maps to Operations on Matrices

Suppose two linear maps have the same domain and codomain. We can add their outputs for each input, or multiply the output of one map by a scalar. The resulting maps have matrix representations too. The key fact is that, when both maps are represented using the same domain basis and the same codomain basis, these operations are reflected directly in the matrices: addition of maps becomes matrix addition, and scalar multiplication of maps becomes scalar multiplication of matrices.

Throughout, let \(\mathbb{F}\) be either \(\mathbb{R}\) or \(\mathbb{C}\), and let \(V\) and \(W\) be finite-dimensional vector spaces over \(\mathbb{F}\). Fix ordered bases \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) for \(V\) and \(\mathcal{C}=(\mathbf{c}_1,\dots,\mathbf{c}_m)\) for \(W\). The coordinate vector \([\mathbf{v}]_{\mathcal{B}}\) lists the coefficients of \(\mathbf{v}\) in the basis \(\mathcal{B}\). For a linear map \(T:V\to W\), its matrix \([T]_{\mathcal{C}\leftarrow\mathcal{B}}\) is the matrix satisfying \([T(\mathbf{v})]_{\mathcal{C}}=[T]_{\mathcal{C}\leftarrow\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}\) for every \(\mathbf{v}\in V\).

These operations are defined on outputs in \(W\), so the maps must share a codomain. To compare their matrices entry by entry, they must also be represented using the same bases \(\mathcal{B}\) and \(\mathcal{C}\). The theorem below states precisely what follows from that shared choice.

Why Coordinates Respect Addition and Scaling

First, coordinates in a fixed basis respect vector addition and scalar multiplication. This follows from the uniqueness of the coefficients in a basis expansion.

Proof. Write \(\mathbf{u}=\sum_{j=1}^{m}\alpha_j\mathbf{c}_j\) and \(\mathbf{w}=\sum_{j=1}^{m}\beta_j\mathbf{c}_j\). Then \(\mathbf{u}+\mathbf{w}=\sum_{j=1}^{m}(\alpha_j+\beta_j)\mathbf{c}_j\), so the coordinate vector of the sum has entries \(\alpha_j+\beta_j\), exactly the entries of \([\mathbf{u}]_{\mathcal{C}}+[\mathbf{w}]_{\mathcal{C}}\). Also, \(a\mathbf{u}=\sum_{j=1}^{m}(a\alpha_j)\mathbf{c}_j\), so its coordinate vector is \(a[\mathbf{u}]_{\mathcal{C}}\). In each case, uniqueness of basis coordinates identifies the displayed coefficients with the coordinates of the vector. \(\square\)

Proof. Set \(\mathbf{A}=[T]_{\mathcal{C}\leftarrow\mathcal{B}}\) and \(\mathbf{D}=[S]_{\mathcal{C}\leftarrow\mathcal{B}}\). For any \(\mathbf{v}\in V\), the definition of the matrix representation and the coordinate proposition give \([ (T+S)(\mathbf{v})]_{\mathcal{C}}=[T(\mathbf{v})+S(\mathbf{v})]_{\mathcal{C}}=[T(\mathbf{v})]_{\mathcal{C}}+[S(\mathbf{v})]_{\mathcal{C}}=(\mathbf{A}+\mathbf{D})[\mathbf{v}]_{\mathcal{B}}\). Thus \(\mathbf{A}+\mathbf{D}\) gives the correct output coordinates for \(T+S\). For scalar multiplication, \([(aT)(\mathbf{v})]_{\mathcal{C}}=[aT(\mathbf{v})]_{\mathcal{C}}=a[T(\mathbf{v})]_{\mathcal{C}}=a\mathbf{A}[\mathbf{v}]_{\mathcal{B}}\), so \(a\mathbf{A}\) gives the correct output coordinates for \(aT\). To see that these actions determine the matrices, each column of a representing matrix is its action on one standard coordinate vector: choosing \(\mathbf{v}=\mathbf{b}_j\) gives \([\mathbf{v}]_{\mathcal{B}}=\mathbf{e}_j\). Therefore the matrices with these actions have the required columns and are the representing matrices. This argument also covers \(a=0\), when every output coordinate is zero. \(\square\)

Computing with the Shared Bases

The matrices in that example are rectangular, but the rule is unchanged: both maps send the same two-dimensional domain to the same three-dimensional codomain, and both matrices use the same coordinate bases. The common dimensions ensure that matrix addition and subtraction are defined.

The Same Rule in Nonstandard Bases

The theorem does not require standard coordinates. It does require that the chosen bases be shared by the maps being combined. The next example uses a nonstandard basis for both the domain and the codomain, so the coordinate arithmetic remains consistent.

Common Pitfall: Combining Matrices from Different Bases

Matrix addition requires matrices of the same shape, but matching shapes alone do not justify adding representation matrices. Their entries must describe maps using the same domain and codomain coordinate conventions. For example, take the identity map on \(\mathbb{R}^2\). With the standard basis for both domain and codomain, its matrix is \(\mathbf{I}_2\). If its domain basis is instead \(\mathcal{B}=((1,1),(0,1))\) and its codomain basis remains standard, its matrix has columns \((1,1)^{\top}\) and \((0,1)^{\top}\):

A tempting but incorrect calculation is to add this matrix to \(\mathbf{I}_2\) and claim that the result represents twice the identity. The sum would be \(\begin{bmatrix}2&0\\1&2\end{bmatrix}\). But if both copies of the identity use standard bases, twice the identity has matrix \(\begin{bmatrix}2&0\\0&2\end{bmatrix}\). If both copies use domain basis \(\mathcal{B}\) and standard codomain basis, its matrix is instead \(2\begin{bmatrix}1&0\\1&1\end{bmatrix}=\begin{bmatrix}2&0\\2&2\end{bmatrix}\). The incorrect sum mixes two different domain bases. Use the theorem only after confirming that both representation matrices use the same domain basis and the same codomain basis.

The practical consequence is a reliable calculation rule: determine the common domain and codomain bases first, form both representation matrices in those bases, and only then add or scale their entries. The matrix operation mirrors the operation on maps because the output coordinates of a sum or scalar multiple are obtained by the same operation on the individual output coordinates.