- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.14 Isomorphism between linear transformations and matrices
$\quad$ Let $V$ and $W$ be finite-dimensional linear spaces with $\dim V = n$ and $\dim W = m.$ Choose a basis $(e_1, \dots, e_n)$ for $V$ and $(w_1, \dots, w_m)$ for $W.$ In this discussion, these bases will be kept fixed. Let $\mathscr{L}(V,W)$ denote the linear space of all linear transformations of $V$ into $W.$ If $T \in \mathscr{L}(V,W),$ let $m(T)$ denote the matrix of $T$ relative to the given bases. We recall that $m(T)$ is defined as follows.
$\quad$ The image of each basis element $e_k$ is expressed as a linear combination of the basis elements of $W:$ \begin{align*} (2.19) \qquad T(e_k) &= \sum_{i=1}^m t_{ik}w_i \qquad \text{for} \quad k = 1, 2, \dots, n \end{align*} The scalar multipliers $t_{ik}$ are the $ik^{th}$ entries of of $m(T).$ Thus, we have \begin{align*} (2.20) \qquad m(T) &= (t_{ik})_{i,k =1}^{m,n} \end{align*} Equation (2.20) defines a new function $m$ whose domain is the set of linear transformations $\mathscr{L}(V, W)$ and whose values are matrices in $M_{m,n}.$ Since every $m \times n$ matrix is the matrix $m(T)$ for some $T$ in $\mathscr{L}(V,W),$ the range of $m$ is $M_{m,n}.$ In the next theorem, we will prove that the transformation $m: \mathscr{L}(V,W) \to M_{m,n}$ is linear and one-to-one on $\mathscr{L}(V,W).$
$\quad$ Theorem 2.15. $\quad$ Isomorphism Theorem. $\quad$ For all $S$ and $T$ in $\mathscr{L}(V,W)$ and all scalars $c,$ we have \begin{align*} m(S + T) &= m(S) + m(T) \qquad and \qquad m(cT) = cm(T). \end{align*} Moreover, \begin{align*} m(S) &= m(T) \qquad implies \qquad S = T. \end{align*} so $m$ is one-to-one on $\mathscr{L}(V, W).$
$\quad$ Proof. $\quad$ Let $(e_1, \dots, e_n)$ be a basis for $V,$ $(w_1, \dots, w_m)$ be a basis for $W,$ and let $S$ and $T$ be linear transformations in $\mathscr{L}(V,W).$ Then, we can express the image of each $e_k$ under $S$ and $T$ as linear combinations of the basis elements of $W:$ \begin{align*} S(e_k) &= \sum_{i=1}^m s_{ik} w_i, \qquad T(e_k) = \sum_{i=1}^m t_{ik} w_i, \qquad \text{for} \quad k = 1, 2, \dots, n \end{align*} with their respective matrices given by \begin{align*} m(S) &= (s_{ik})_{i, k = 1}^{m,n}, \qquad m(T) = (t_{ik})_{i, k = 1}^{m,n} \end{align*} The sum of these two images is then given by: \begin{align*} (S + T)(e_k) &= \sum_{i=1}^m (s_{ik} + t_{ik}) w_i, \qquad \text{for} \quad k = 1, 2, \dots, n \end{align*} with matrix $m(S + T)$ given by: \begin{align*} m(S + T) &= (s_{ik} + t_{ik})_{i, k = 1}^{m,n} \end{align*} But by definition, the sum of two $m \times n$ matrices $A = (a_{ik})$ and $B = (b_{ik})$ is $A + B = (a_{ik} + b_{ik}).$ As such, this means that $m(S + T) = m(S) + m(T).$
$\quad$ Similarly, if $c$ is a scalar, we have \begin{align*} cS(e_k) &= \sum_{i=1}^m cs_{ik} w_i, \qquad \text{for} \quad k = 1, 2, \dots, n \end{align*} with matrix $m(cS)$ given by \begin{align*} m(cS) &= (cs_{ik})_{i, k = 1}^{m,n} \end{align*} then, by definition, $m(cS) = cm(S).$ Thus, we have shown that $m(S + T) = m(S) + m(T)$ and $m(cT) = cm(T)$ for transformations $S$ and $T$ in $\mathscr{L}(V,W)$ and any scalar $c.$
$\quad$ Now, to show that $m$ is one-to-one on $\mathscr{L}(V, W),$ let $S$ and $T$ be linear transformations in $\mathscr{L}(V,W)$ such that $m(S) = m(T).$ If $m(S) = (s_{ik})$ and $m(T) = (t_{ik}),$ then by definition, $s_{ik} = t_{ik}$ for each pair $(i, k).$ Moreover, if $(e_1, \dots, e_n)$ is a basis for $V,$ the image of each basis element $e_k$ is given by \begin{align*} S(e_k) &= \sum_{i=1}^m s_{ik} w_i, \qquad T(e_k) = \sum_{i=1}^m t_{ik} w_i, \qquad \text{for} \quad k = 1, 2, \dots, n \end{align*} where $(w_1, \dots, w_m)$ is a basis for $W.$ But if $s_{ik} = t_{ik}$ for each pair $(i, k),$ it follows that $S(e_k) = T(e_k)$ for $k = 1, \dots, n.$ Then, if we let $x = c_1e_1 + \dots + c_ne_n$ be an arbitrary element of $V,$ we have: \begin{align*} S(x) &= \sum_{k=1}^n c_kS(e_k) \\ &= \sum_{k=1}^n c_kT(e_k) \\ &= T\left(\sum_{k=1}^n c_ke_k \right) \\ &= T(x) \end{align*} But since this is true for any $x$ in $V,$ it follows that $S = T.$ Thus, we have shown that $m$ is one-to-one on $\mathscr{L}(V, W),$ completing the proof.
$\quad$ Note: $\quad$ The function $m$ is called an isomorphism. For a given choice of bases, $m$ establishes a one-to-one correspondence between the set of linear transformations $\mathscr{L}(V,W)$ and the set of $m \times n$ matrices $M_{m,n}.$ The operations of addition and scalar multiplication are preserved under this correspondence. The linear spaces $\mathscr{L}(V,W)$ and $M_{m,n}$ are said to be isomorphic. Incidentally, Theorem 2.11 shows that the domain of a one-to-one linear transformation has the same dimension as its range. Therefore, $\dim \mathscr{L}(V, W) = \dim M_{m,n} = mn.$