linear transformations and derivatives(FUNCTIONS OF SERVERAL VARIABLES)

Linear transformations A mapping $L$ from a vector space $V$ into a vector space $W$ is said to be a linear transformation if

$$ \begin{equation} \begin{aligned} L(\alpha \textbf{v}_{1}+\beta \textbf{v}_{2})= \alpha L(\textbf{v}_{1})+\beta L(\textbf{v}_{2}) \end{aligned}\end{equation} $$

for all $\textbf{v}_{1} ,\textbf{v}_{2} \in V$ and for all scalars $\alpha$ and $\beta$.

  In fact, for every linear transformation there exists a matrix corresponding to it.

  Consider a linear transformation $L$ from $V$ into $W$, where $V$ and $W$ are vector spaces of dimension $n$ and $m$, respectively. Let $E=\{\textbf{v}_{1} ,\textbf{v}_{2}, \dots, \textbf{v}_{n}\}$ be an ordered basis for $V$ and $F=\{\textbf{w}_{1} ,\textbf{w}_{2}, \dots, \textbf{w}_{m}\}$ be an ordered basis for $W$. Let $L$ be a linear transformation mapping $V$ into $W$. If $\textbf{v}$ is any vector in $V$, then we can express $\textbf{v}$ in terms of the basis $E$:

$$ \begin{equation} \begin{aligned} \textbf{v}=\sum_{j=1}^{n}x_{j} \textbf{v}_{j}. \end{aligned}\end{equation} $$

  For $j=1,2,\dots,n$, let $\textbf{a}_{j}=(a_{1j},a_{2j},\dots,a_{mj})^{T}$ be the coordinate vector of $L(\textbf{v}_{j})$ with respect to $\{\textbf{w}_{1} ,\textbf{w}_{2}, \dots, \textbf{w}_{m}\}$; i.e.,

$$ \begin{equation} \begin{aligned} L(\textbf{v}_{j})=\sum_{k=1}^{m}a_{kj} \textbf{w}_{k}. \end{aligned}\end{equation} $$

Then,

$$ \begin{equation} \begin{aligned} L(\textbf{v})&=L(\sum_{j=1}^{n}x_{j} \textbf{v}_{j}) \\ &=\sum_{j=1}^{n}x_{j}L(\textbf{v}_{j}) \\ &=\sum_{j=1}^{n}x_{j}(\sum_{k=1}^{m}a_{kj} \textbf{w}_{k}) \\ &=\sum_{k=1}^{m}(\sum_{j=1}^{n}x_{j}a_{kj}) \textbf{w}_{k}. \end{aligned}\end{equation} $$

  If we let $A=(a_{ij})=(\textbf{a}_{1},\textbf{a}_{2},\dots,\textbf{a}_{n})$ and $y_{i}=\sum_{j=1}^{n}x_{j}a_{kj}$, then

$$ \begin{equation} \begin{aligned} \textbf{y}_{i}=(y_{1},y_{2},\dots,y_{m})^{T}=A\textbf{x} \end{aligned}\end{equation} $$

is the coordinate vector of $L(\textbf{v})$ with respect to $\{\textbf{w}_{1} ,\textbf{w}_{2}, \dots, \textbf{w}_{m}\}$. We therefore find a corresponding matrix $A$ to $L(V,W)$.

Derivatives

  If $f$ is a real function with domain $(a,b) \subset R^{1}$ and if $x\in (a,b)$, then $f'(x)$ is usually defined to be the real number

$$ \begin{equation} \begin{aligned} \lim_{h\rightarrow0} \frac{f(x+h)-f(x)}{h}, \end{aligned}\end{equation} $$

provided, of course, that the limit exists. Thus

$$ \begin{equation} \begin{aligned} f(x+h)-f(x)=f'(x)h+r(h) \end{aligned}\end{equation} $$

where the “remainder” $r(h)$ is small, in the sense that

$$ \begin{equation} \begin{aligned} \lim_{h\rightarrow0} \frac{r(h)}{h}=0. \end{aligned}\end{equation} $$

  Note that the equality above might indicate that the difference $f(x+h)-f(x)$ could be approximated by a linear function $f'(x)h$. The point is, we could also regard the derivative of $f$, this linear function, as a linear operator on $R^{1}$ which takes $h$ to $f'(x)h$.

  Then we can naturally extend the concept of derivative to a higher dimensional space. In this case the derivative of function $\textbf{f}$ should be a linear transformation, or, equivalently, a matrix.

9.11 Definition Suppose $E$ is an open set in $R^{n}$, $\textbf{f}$ maps $E$ into $R^{m}$, and $\textbf{x}\in E$. If there exists a linear transformation $A$ of $R^{n}$ into $R^{m}$ such that

$$ \begin{equation} \begin{aligned} \lim_{\textbf{h}\rightarrow0} \frac{\vert \textbf{f}(\textbf{x}+\textbf{h})- \textbf{f}(\textbf{x})- A\textbf{h}\vert} {\vert\textbf{h}\vert}=\textbf{0}, \end{aligned}\end{equation} $$

then we say that $\textbf{f}$ is differentiable at $\textbf{x}$, and we write

$$ \begin{equation} \begin{aligned} \textbf{f}'(\textbf{x})=A. \end{aligned}\end{equation} $$