Linear Maps from Fn to Fm

Recall that the vector space $\mathbb{F}^n$ is the set of all vectors with $n$ components where the values of the components are from $\mathbb{F}$. Similarly, $\mathbb{F}^m$ is the set all vectors with $m$ components where the values of the components are form $\mathbb{F}$.

Also recall from the Linear Maps Defined by Bases page that if $\{ v_1, v_2, ..., v_n \}$ is a basis of $V$ and if $\{ w_1, w_2, ..., w_n \}$ is any set of vectors in $W$ then we can define a unique linear map $T : V \to W$ by $T(v_j) = w_j$ for each $j = 1, 2, ..., n$.

In the following theorem, we will show that every linear map $T \in \mathcal L (\mathbb{F}^n, \mathbb{F}^m)$ for $A_{j,k} \in \mathbb{F}$ for each $j = 1, 2, ..., m$ and for each $k = 1, 2, ..., n$ and for all $(x_1, x_2, ..., x_n) \in \mathbb{F}^n$ is of the form:

(1)

\begin{align} \quad T(x_1, x_2, ..., x_n) = (A_{1,1}x_1 + A_{1,2}x_2 + ... + A_{1,n}x_n, A_{2,1}x_1 + A_{2,2}x_2 + ... + A_{2,n}x_n, ..., A_{m,1}x_1 + A_{m,2}x_2 + ... + A_{m,n}x_n) \end{align}

Theorem 1: Let $T \in \mathcal L (\mathbb{F}^n, \mathbb{F}^m)$. Then there exists scalars $A_{j,k}$ for $j = 1, 2, ..., m$ and for $k = 1, 2, ..., n$ such that for every vector $(x_1, x_2, ..., x_n) \in \mathbb{F}^n$ we have that $T(x_1, x_2, ..., x_n) = (A_{1,1}x_1 + A_{1,2}x_2 + ... + A_{1,n}x_n, A_{2,1}x_1 + A_{2,2}x_2 + ... + A_{2,n}x_n, ..., A_{m,1}x_1 + A_{m,2}x_2 + ... + A_{m,n}x_n)$.

• Proof: We note that $\mathbb{F}^n$ is a finite-dimensional vector space. Consider the standard basis of $\mathbb{F}^n$:

(2)

\begin{align} \quad \{ (1, 0, 0, ..., 0), (0, 1, 0, ... 0), ..., (0, 0, 0, ..., 1) \} = \{ e_1, e_2, ..., e_n \} \end{align}

• For scalars $A_{j,k} \in \mathbb{F}$ for $j = 1, 2, ..., m$ and for $k = 1, 2, ..., n$ define the linear map $T$ by:

(3)

\begin{align} \quad T(1, 0, 0, ..., 0) = T(e_1) = (A_{1,1}, A_{2, 1}, ..., A_{m,1}) \\ \quad T(0, 1, 0, ..., 0) = T(e_2) = (A_{1,2}, A_{2,2}, ..., A_{m, 2}) \\ \quad \quad \quad \vdots \quad \quad \quad \\ \quad T(0, 0, 0, ..., 1) = T(e_n) = (A_{1,n}, A_{2,n}, ..., A_{m,n}) \end{align}

• By the theorem on the Linear Maps Defined by Bases we have that since $\{ (1, 0, 0, ..., 0), (0, 1, 0, ... 0), ..., (0, 0, 0, ..., 1) \} = \{ e_1, e_2, ..., e_n \}$ is a basis of $\mathbb{F}^n$ and since $\{ (A_{1,1}, A_{2, 1}, ..., A_{m,1}), (A_{1,2}, A_{2,2}, ..., A_{m, 2}), ..., (A_{1,n}, A_{2,n}, ..., A_{m,n}) \}$ is a set of vectors in $\mathbb{F}^m$, then $T$ is the unique linear map $T \in \mathcal ( \mathbb{F}^n, \mathbb{F}^m)$ defined by above, and so:

(4)

\begin{align} \quad T(x_1, x_2, ..., x_n) = T(x_1e_1 + x_2e_2 + ... + x_ne_n) = x_1T(e_1) + x_2T(e_2) + ... + x_nT(e_n) \\ \quad = x_1(A_{1,1}, A_{2, 1}, ..., A_{m,1}) + x_2(A_{1,2}, A_{2,2}, ..., A_{m, 2}) + ... + x_n (A_{1,n}, A_{2,n}, ..., A_{m,n}) \\ \quad = (A_{1,1}x_1, A_{2, 1}x_1, ..., A_{m,1}x_1) + (A_{1,2}x_2, A_{2,2}x_2, ..., A_{m, 2}x_2) + ... + (A_{1,n}x_n, A_{2,n}x_n, ..., A_{m,n}x_n) \\ \quad = (A_{1,1}x_1 + A_{1,2}x_2 + ... + A_{1,n}x_n, A_{2,1}x_1 + A_{2,2}x_2 + ... + A_{2,n}x_n, ..., A_{m,1}x_1 + A_{m,2}x_2 + ... + A_{m,n}x_n) \quad \blacksquare \end{align}