Matrix Units as a Basis for Linear Maps

Building Maps One Basis Vector at a Time

A linear map is determined by what it does to a basis of its domain. This gives a direct way to build a basis for the space of all linear maps: make one map for each pair consisting of a domain basis vector and a codomain basis vector. Each such map sends its selected domain vector to its selected codomain vector and sends every other domain basis vector to zero. These maps are the matrix units for linear maps.

Let \(V\) and \(W\) be finite-dimensional vector spaces over a field \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\). Fix ordered bases \(\mathcal{B}=(\mathbf{v}_1,\dots,\mathbf{v}_n)\) for \(V\) and \(\mathcal{C}=(\mathbf{w}_1,\dots,\mathbf{w}_m)\) for \(W\). The space \(L(V,W)\) consists of all linear maps from \(V\) to \(W\). We use the usual pointwise addition and scalar multiplication of maps.

The two indices have distinct roles. The first index \(i\) selects the output basis vector in \(W\); the second index \(j\) selects the input basis vector in \(V\). In coordinates relative to \(\mathcal{B}\) and \(\mathcal{C}\), the matrix of \(\mathcal{E}_{ij}\) has a \(1\) in row \(i\), column \(j\), and zeros elsewhere. Thus these maps generalize the single-entry matrices often called matrix units.

Every Linear Map Has a Unique Expansion

To find the coefficients of a map \(T\in L(V,W)\), evaluate it on each domain basis vector. Since \(\mathcal{C}\) is a basis, each value \(T(\mathbf{v}_j)\) has a unique expansion in the vectors \(\mathbf{w}_1,\dots,\mathbf{w}_m\). The coefficient of \(\mathbf{w}_i\) in that expansion is exactly the coefficient of \(\mathcal{E}_{ij}\) in the expansion of \(T\).

Proof. Let \(T\in L(V,W)\). For each \(j\), the vector \(T(\mathbf{v}_j)\in W\) has unique coordinates \(a_{1j},\dots,a_{mj}\) in the basis \(\mathcal{C}\), so \(T(\mathbf{v}_j)=\sum_{i=1}^{m}a_{ij}\mathbf{w}_i\). Define \(S=\sum_{j=1}^{n}\sum_{i=1}^{m}a_{ij}\mathcal{E}_{ij}\). For a fixed \(k\), the definition of the maps gives \(\mathcal{E}_{ij}(\mathbf{v}_k)=\mathbf{0}\) unless \(j=k\), in which case it gives \(\mathbf{w}_i\). Therefore \(S(\mathbf{v}_k)=\sum_{i=1}^{m}a_{ik}\mathbf{w}_i=T(\mathbf{v}_k)\). Every \(\mathbf{v}\in V\) is a linear combination of the \(\mathbf{v}_k\), so linearity gives \(S(\mathbf{v})=T(\mathbf{v})\). Hence the maps span \(L(V,W)\). To prove independence, suppose \(\sum_{j=1}^{n}\sum_{i=1}^{m}c_{ij}\mathcal{E}_{ij}\) is the zero map. Fix \(k\) and apply this map to \(\mathbf{v}_k\). All terms with \(j\ne k\) vanish, leaving \(\sum_{i=1}^{m}c_{ik}\mathbf{w}_i=\mathbf{0}\). The vectors in \(\mathcal{C}\) are linearly independent, so \(c_{ik}=0\) for every \(i\). This holds for each \(k\), and thus every coefficient \(c_{ij}\) is zero. The maps are linearly independent as well as spanning, so they form a basis. The same argument shows that the coefficients in the displayed expansion must be the coordinates of each \(T(\mathbf{v}_j)\), and hence are unique. \(\square\)

This proof also covers zero-dimensional cases. If \(V\) or \(W\) is the zero space, there is only the zero map in \(L(V,W)\), and there are no index pairs \((i,j)\). The resulting empty list spans the zero space and is linearly independent, so it is a basis under the usual convention for the zero space.

Proof. If \(T\) and \(S\) agree on every \(\mathbf{v}_j\), take any \(\mathbf{v}\in V\) and write \(\mathbf{v}=\sum_{j=1}^{n}c_j\mathbf{v}_j\). Linearity gives \(T(\mathbf{v})=\sum_{j=1}^{n}c_jT(\mathbf{v}_j)=\sum_{j=1}^{n}c_jS(\mathbf{v}_j)=S(\mathbf{v})\). Thus \(T=S\). For the coefficient statement, expand each \(T(\mathbf{v}_j)\) in the basis \(\mathcal{C}\); the theorem identifies its coefficient of \(\mathbf{w}_i\) with the coefficient of \(\mathcal{E}_{ij}\). \(\square\)

Worked Examples

Why the Basis Depends on Both Choices

The map units are built from the selected bases, so changing either basis changes the individual maps. Their defining roles do not change: one domain basis vector is selected, one codomain basis vector is selected, and all other domain basis vectors are sent to zero. The coefficient array is precisely the matrix of the map relative to the chosen domain and codomain bases. In particular, the row index refers to the codomain coordinate and the column index refers to the domain basis vector.

This construction is useful whenever the task is to describe a linear map by its action on basis vectors. Instead of trying to guess a formula for the whole map, specify one output for each domain basis vector, express each output in the codomain basis, and read off the coefficients. The resulting matrix-unit expansion both records those values and verifies that they determine one and only one linear map.

Common Indexing Pitfall

A frequent mistake is to reverse the indices and treat \(\mathcal{E}_{ij}\) as the map sending \(\mathbf{v}_i\) to \(\mathbf{w}_j\). That reverses the roles of domain and codomain. For example, with \(V=\mathbb{R}^2\), \(W=\mathbb{R}^3\), and standard bases, \(\mathcal{E}_{21}\) sends the first domain vector to the second codomain vector. Its matrix is

The incorrect reversed interpretation would place its \(1\) in row \(1\), column \(2\), which would give a \(3\times2\) matrix with the wrong action: it would send the second domain basis vector to the first codomain basis vector. Check the definition before placing the nonzero entry: the column identifies the selected input, and the row identifies the output.