Why the Number of Maps Is Controlled by Basis Values
A linear map is determined by what it does to a basis of its domain. If the domain basis has \(n\) vectors, then specifying a map means specifying \(n\) output vectors in the codomain. When the codomain has dimension \(m\), each output vector requires \(m\) scalar coordinates. This suggests \(mn\) scalar choices. The dimension formula makes that count precise: these choices are independent, and together they give coordinates for every linear map.
Throughout this tutorial, \(V\) and \(W\) are finite-dimensional vector spaces over \(\mathbb{F}\), where \(\mathbb{F}\) is either \(\mathbb{R}\) or \(\mathbb{C}\). Write \(\dim V=n\) and \(\dim W=m\). The notation \(L(V,W)\) means the vector space of all linear maps from \(V\) to \(W\), with addition and scalar multiplication defined pointwise. A basis of a vector space is an ordered list that is linearly independent and spans that space. Its length is the dimension of the space.
The product counts one scalar for each choice of a domain basis vector and a codomain coordinate. To establish that count as a dimension, we must show both that these scalars determine every map and that no different choices describe the same map.
Values on a Basis Determine a Linear Map
Proof. If \(S=T\), then they have the same value at every vector, including each \(\mathbf{b}_j\). Conversely, suppose their values agree on every basis vector. Each \(\mathbf{v}\in V\) has a unique expression \(\mathbf{v}=\sum_{j=1}^n x_j\mathbf{b}_j\), with scalars \(x_j\). By linearity, \(S(\mathbf{v})=\sum_{j=1}^n x_jS(\mathbf{b}_j)=\sum_{j=1}^n x_jT(\mathbf{b}_j)=T(\mathbf{v})\). Thus \(S\) and \(T\) agree on every vector of \(V\), so \(S=T\). \(\square\)
There is also no compatibility condition tying together the images of different basis vectors. Any chosen list \(\mathbf{w}_1,\dots,\mathbf{w}_n\) in \(W\) is the list of basis values of exactly one linear map.
Proof. Every \(\mathbf{v}\in V\) has a unique basis expansion \(\mathbf{v}=\sum_{j=1}^n x_j\mathbf{b}_j\). Define \(T(\mathbf{v})=\sum_{j=1}^n x_j\mathbf{w}_j\). This definition gives one output for each \(\mathbf{v}\), because the coefficients \(x_j\) are unique. If \(\mathbf{u}=\sum_j y_j\mathbf{b}_j\) and \(a\in\mathbb{F}\), then \(T(\mathbf{u}+\mathbf{v})=\sum_j(y_j+x_j)\mathbf{w}_j=T(\mathbf{u})+T(\mathbf{v})\) and \(T(a\mathbf{v})=\sum_j(ax_j)\mathbf{w}_j=aT(\mathbf{v})\). Therefore \(T\) is linear. Substituting \(\mathbf{v}=\mathbf{b}_k\) gives \(T(\mathbf{b}_k)=\mathbf{w}_k\). Finally, any other linear map with those basis values agrees with \(T\) on every basis vector and hence equals \(T\) by the lemma. \(\square\)
Coordinate Maps and the Dimension Proof
Choose a basis \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) of \(V\) and a basis \(\mathcal{C}=(\mathbf{c}_1,\dots,\mathbf{c}_m)\) of \(W\). For each pair of indices \(i,j\), define a map \(\mathbf{E}_{ij}:V\to W\) by specifying its values on \(\mathcal{B}\): it sends \(\mathbf{b}_j\) to \(\mathbf{c}_i\), and it sends each other basis vector to \(\mathbf{0}\). The proposition guarantees that this specifies exactly one linear map. These maps isolate one output coordinate for one input basis vector.
To prove that they span, take any \(T\in L(V,W)\). For each \(j\), express its value in the codomain basis: $$ T(\mathbf{b}_j)=\sum_{i=1}^m a_{ij}\mathbf{c}_i. $$ Then form the map \(S=\sum_{j=1}^n\sum_{i=1}^m a_{ij}\mathbf{E}_{ij}\). For any \(k\), the definition of the coordinate maps gives $$ S(\mathbf{b}_k)=\sum_{j=1}^n\sum_{i=1}^m a_{ij}\mathbf{E}_{ij}(\mathbf{b}_k) =\sum_{i=1}^m a_{ik}\mathbf{c}_i =T(\mathbf{b}_k). $$ The maps \(S\) and \(T\) agree on a basis, so the lemma gives \(S=T\). Thus the coordinate maps span \(L(V,W)\).
To prove linear independence, suppose scalars \(d_{ij}\) satisfy \(\sum_{j=1}^n\sum_{i=1}^m d_{ij}\mathbf{E}_{ij}=\mathbf{0}\), where \(\mathbf{0}\) here denotes the zero map. Evaluate both sides at \(\mathbf{b}_k\). The result is $$ \sum_{i=1}^m d_{ik}\mathbf{c}_i=\mathbf{0}. $$ The vectors \(\mathbf{c}_1,\dots,\mathbf{c}_m\) are linearly independent, so \(d_{ik}=0\) for every \(i\). This holds for every \(k\), hence every coefficient \(d_{ij}\) is zero. The maps are linearly independent as well as spanning, so they form a basis. There are \(m\) choices for \(i\) and \(n\) choices for \(j\), giving \(mn\) basis maps and therefore \(\dim L(V,W)=mn\).
It remains to handle a zero dimension. If \(V\) is the zero space, the only map \(V\to W\) is the zero map. If \(W\) is the zero space, every map \(V\to W\) must also be the zero map. In either case \(L(V,W)\) consists of just its zero vector, so its dimension is \(0\), agreeing with \(mn\) when \(m=0\) or \(n=0\). This completes the proof for all finite dimensions. \(\square\)
Reading the Count in Coordinates
Worked Example: Maps from \(\mathbb{R}^2\) to \(\mathbb{R}^3\)
Use the standard bases \((\mathbf{e}_1,\mathbf{e}_2)\) of \(\mathbb{R}^2\) and \((\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3)\) of \(\mathbb{R}^3\). Write the images of the two domain basis vectors using six scalars: $$ T(\mathbf{e}_1)= \begin{bmatrix}a_{11}\\a_{21}\\a_{31}\end{bmatrix}, \qquad T(\mathbf{e}_2)= \begin{bmatrix}a_{12}\\a_{22}\\a_{32}\end{bmatrix}. $$
For \(\mathbf{x}=\begin{bmatrix}x_1\\x_2\end{bmatrix}=x_1\mathbf{e}_1+x_2\mathbf{e}_2\), linearity gives $$ T(\mathbf{x})= \begin{bmatrix} a_{11}x_1+a_{12}x_2\\ a_{21}x_1+a_{22}x_2\\ a_{31}x_1+a_{32}x_2 \end{bmatrix}. $$ There are three coordinates for each of two basis images, so there are six scalar parameters. Equivalently, the six coordinate maps \(\mathbf{E}_{ij}\), with \(1\leq i\leq3\) and \(1\leq j\leq2\), form a basis. The dimension formula gives \(3\cdot2=6\).
Check: Substituting \(\mathbf{x}=\mathbf{e}_1\) gives \(\begin{bmatrix}a_{11}\\a_{21}\\a_{31}\end{bmatrix}=T(\mathbf{e}_1)\). Substituting \(\mathbf{x}=\mathbf{e}_2\) gives \(\begin{bmatrix}a_{12}\\a_{22}\\a_{32}\end{bmatrix}=T(\mathbf{e}_2)\). Thus the formula recovers the specified basis values, which determine the map uniquely.
Worked Example: A Map from Polynomials to \(\mathbb{R}^2\)
Let \(V=P_2(\mathbb{R})\), the space of real polynomials of degree at most \(2\), with basis \((1,x,x^2)\), and let \(W=\mathbb{R}^2\) with its standard basis. Specify a linear map by \(T(1)=\begin{bmatrix}1\\1\end{bmatrix}\), \(T(x)=\begin{bmatrix}2\\-1\end{bmatrix}\), and \(T(x^2)=\begin{bmatrix}3\\4\end{bmatrix}\).
For \(p(x)=a+bx+cx^2\), linearity and the given basis values yield $$ T(p)=aT(1)+bT(x)+cT(x^2) =\begin{bmatrix}a+2b+3c\\a-b+4c\end{bmatrix}. $$ Thus the two output coordinates for each of the three domain basis vectors give six scalar coordinates for a general map. Here the chosen map has coefficient pairs \((1,1)\), \((2,-1)\), and \((3,4)\), one pair for each basis vector. The formula also shows how every possible selection of three output vectors defines a map.
Check: For \(p(x)=1\), the formula gives \(\begin{bmatrix}1\\1\end{bmatrix}\). For \(p(x)=x\), it gives \(\begin{bmatrix}2\\-1\end{bmatrix}\). For \(p(x)=x^2\), it gives \(\begin{bmatrix}3\\4\end{bmatrix}\). These are exactly the specified values, so the formula agrees with the definition on a basis.
Worked Example: A Zero-Dimensional Codomain
Let \(V=\mathbb{R}^2\) and let \(W=\{\mathbf{0}\}\), the zero vector space. Any function \(T:V\to W\) must send every input to the only available output, \(\mathbf{0}\). In particular, \(T(\mathbf{e}_1)=\mathbf{0}\) and \(T(\mathbf{e}_2)=\mathbf{0}\); by linearity, \(T(x_1\mathbf{e}_1+x_2\mathbf{e}_2)=x_1\mathbf{0}+x_2\mathbf{0}=\mathbf{0}\). There is exactly one linear map, the zero map.
The space \(L(V,W)\) therefore has no nonzero vectors and has dimension \(0\). The formula agrees: \((\dim V)(\dim W)=2\cdot0=0\). There are no coordinate maps \(\mathbf{E}_{ij}\) to list because the codomain basis is empty; the empty list is a basis of the zero vector space.
Check: The map just described sends every vector in \(V\) to \(\mathbf{0}\), so it is the zero map. Any other map into \(W\) would have to take the same sole possible value at every input and hence would equal it.
Common Mistake: Adding Instead of Multiplying
A frequent error is to count the basis values in the domain and codomain and add their dimensions. For maps from \(\mathbb{R}^2\) to \(\mathbb{R}^3\), that incorrect calculation gives \(2+3=5\). But the map requires two output vectors, each with three coordinates: six scalar choices in total. In the first worked example those choices are \(a_{11},a_{21},a_{31},a_{12},a_{22},a_{32}\), so the correct dimension is \(2\cdot3=6\).
Another way to expose the mistake is to compare maps that differ in just one coordinate. Changing \(a_{21}\) changes the second coordinate of \(T(\mathbf{e}_1)\) without changing the other five parameters. Each parameter corresponds to one coordinate map \(\mathbf{E}_{ij}\), and the proof showed that these maps are independent. None of the \(mn\) choices can be discarded as redundant.
The finite-dimensional hypothesis matters: the proof counts the members of finite bases of both \(V\) and \(W\). It does not claim a finite dimension for \(L(V,W)\) when either space is infinite-dimensional. In the finite-dimensional setting, the result gives a quick dimension count and clarifies what information determines a map: its values on a domain basis, recorded in codomain coordinates.
Check Your Understanding
Use basis values and codomain coordinates to justify each answer.
- If \(\dim V=4\) and \(\dim W=2\), what is \(\dim L(V,W)\), and how many coordinates specify a map on a basis of \(V\)?
- Explain why two linear maps that agree on a basis of their domain must agree on every vector in the domain.
- For \(\dim V=1\) and \(\dim W=3\), how many coordinate maps \(\mathbf{E}_{ij}\) form a basis of \(L(V,W)\)?
- Why can any three chosen vectors in \(W\) be prescribed as the images of a basis of a three-dimensional domain?
- What is \(\dim L(V,W)\) if \(V\) is the zero space and \(W\) has dimension \(5\)? Explain why.
