The Matrix of Linear Transformation and Linear Functionals

Let $B=\{\alpha_1, \alpha_2, ... , \alpha_n\}$ be an ordered basis for $V$ and $B'=\{\beta_1, \beta_2, ... , \beta_n\}$ be an ordered basis for $W$.
Let $T$ be a linear transformation for $V$ onto $W$. Then $T$ is determined by its action on each vector $\alpha_j$ in $V$.
Each $T(\alpha_j)$ can be expressed as
$T(\alpha_j)=\displaystyle\sum_{i=1}^{m}A_{ij}\beta_i$      $(1\le j \le n)$
where $A_{1j}, A_{2j}, ... , A_{mj}$ are scalars and are the coordinates of $T(\alpha_j)$
The $m\times n$ matrix A defined by $A(i, j)=A_{ij}, (1\le i \le m \ \& \ 1\le j \le n)$ is the matrix of $T$ relative to the basis $B$ and $B'$.
Conversely, if $\alpha \in V$, then
$\alpha=a_1\alpha_1+a_2\alpha_2+ ... +a_n\alpha_n$
$=\displaystyle\sum_{j=1}^{n}a_j\alpha_j$
$\implies T(\alpha)=T\left(\displaystyle\sum_{j=1}^{n}a_j\alpha_j\right)$
$=\displaystyle\sum_{j=1}^{n}a_jT(\alpha_j)$
$=\displaystyle\sum_{j=1}^{n}\displaystyle\sum_{i=1}^{m}a_jA_{ij}\beta_i$
$\implies T(\alpha)=\displaystyle\sum_{j=1}^{n}\displaystyle\sum_{i=1}^{m}a_jA_{ij}\beta_i$                 ....... (1)
If $A$ is $m \times n$ matrix on the field $F$, then the definition of $T$ is as given in (1).

Definition:

Let $V$ be a vector space over the field $F$. A linear transformation from $V$ into $V$ is called a linear operator defined on $V$.

Example:

Let $F$ be a field and $T$ be a linear operator defined on $F^2$ as $T(x_1, x_2)=(x_1, 0)$. Then find the matrix $T$ w.r.t. the standard ordered basis for $F^2$.

Solution:

Let $B=\{e_1, e_2\}$ be the standard basis for the $F^2$ where $e_1=(1, 0), e_2=(0,1)$.
Then,
$T(e_1)=T(1, 0)=(1, 0)=1.e_1+0.e_2$
$T(e_2)=T(0, 1)=(0, 0)=0.e_1+0.e_2$
Therefore, the matrix of $T$ w.r.t. $B$ is $[T]_B=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$

Example:

Let $V$ the space of all polynomial functions defined on $\mathbb{R}$ of the type $f(x)=a_0+a_1x+a_2x^2+a_3x^3$. Let $D$ be the differential operator which is defined on $V$. Let $B=\{f_1, f_2, f_3, f_4\}$ be an ordered basis such that $f_j(x)=x^{j-1}$ then find $[D]_B$.

Solution:

$D(f_1(x))=D(x^0)=D(1)=0=0.f_1+0.f_2+0.f_3+0.f_4$
$D(f_2(x))=D(x^1)=D(x)=1=1.f_1+0.f_2+0.f_3+0.f_4$
$D(f_3(x))=D(x^2)=2x=0.f_1+2.f_2+0.f_3+0.f_4$
$D(f_4(x))=D(x^3)=3x^2=0.f_1+0.f_2+3.f_3+0.f_4$
$[D]_B=\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

Definition:

Let $A$ and $B$ be $m \times n$ matrices over the field $F$. Then $B$ is said to be similar to $A$ if there exists an invertible matrix $P$ such that $B=P^{-1}AP$.

Linear Functionals:

Let $V$ be a vector space over the field $F$. Then a linear transformation $f$ from $V$ into $F$ is called a linear functionals. i.e. $f:V \rightarrow F$ is called a linear functional if for any vector $\alpha, \beta \in V \ \text{and} \ c \in F$,
$f(c\alpha+\beta)=cf(\alpha)+f(\beta)$

Example:

Let $F$ be a field and for some scalars $a_1, a_2, ... , a_n$ define $f$ on $F^n$ by $f(x_1, x_2, ... , x_n)=a_1x_1+a_2x_2+ ... +a_nx_n$. Then show that $f$ is a linear functional on $F^n$.

Solution:

Let $x, y \in F^n$
$\therefore x=(x_1, x_2, ..., x_n)\ \& \ y=(y_1, y_2, ..., y_n)$
Let $c \in F$
Consider,
$f(cx+y)$
$=f(c(x_1, x_2, ..., x_n)+(y_1, y_2, ..., y_n))$
$=f((cx_1, cx_2, ..., cx_n)+(y_1, y_2, ..., y_n))$
$=f(cx_1+y_1, cx_2+y_2 ..., cx_n+y_n)$
$=a_1(cx_1+y_1)+a_2(cx_2+y_2)+ ... +a_n(cx_n+y_n)$
$=ca_1x_1+a_1y_1+ca_2x_2+a_2y_2+ ... +ca_nx_n+a_ny_n$
$=(ca_1x_1+ca_2x_2+...+ca_nx_n)+(a_1y_1+a_2y_2+ ... +a_ny_n)$
$=c(a_1x_1+a_2x_2+...+a_nx_n)+(a_1y_1+a_2y_2+ ... +a_ny_n)$
$=cf(x_1, x_2, ..., x_n)+f(y_1, y_2, ..., y_n)$
$\therefore f(cx+y)=cf(x)+f(y)$
$\therefore f$ is linear functional.

Note:

The matrix of $f$ w.r.t. the standard basis $B=\{e_1, e_2, ... , e_n\}$ (where $e_i=(0, 0, ... , 1_{i^{th}}, ... , 0)$ and $B'=\{1\}$ is given by $[a_1, a_2, ... , a_n]$.
Here $f(e_j)=a_j$, for each $j \ (1\le j\le n)$
Now, $f(x_1, x_2, ..., x_n)=f\left(\displaystyle\sum_{j=1}^{n}x_je_j\right)$
$=\displaystyle\sum_{j=1}^{n}x_jf(e_j)$
$=\displaystyle\sum_{j=1}^{n}a_jx_j$

Definition:

Let $A$ be $n\times n$ matrix over the filed F, i.e. $A=(A_{ij})$, then $trace(A)=A_{11}+A_{22}+ ... +A_{nn}$.

Example:

Define $tr:F^{n\times n}\rightarrow F$ by $tr(A)=A_{11}+A_{22}+ ... +A_{nn}$ ($F^{n\times n}$ is the set of all $n\times n$ matrices in $F$). Then tr is linear functional.

Solution:

Let $A, B \in F^{n\times n}$ and $c\in F$.
$tr(cA+B)=cA_{11}+cA_{22}+ ... +cA_{nn}+B_{11}+B_{22}+ ... +B_{nn}$
$=\displaystyle\sum_{i=1}^{n}\left(cA_{ii}+B_{ii}\right)$
$=c\displaystyle\sum_{i=1}^{n}A_{ii}+\displaystyle\sum_{i=1}^{n}B_{ii}$
$=c \ tr(A)+tr(B)$

Note:

If $V$ is a vector space over the field $F$, the set of all linear functionals $f:V\rightarrow F$ is denoted by $L(V, F)$ and $L(V, F)$ is a vector space over the field $F$. w.r.t. following
(i) $(f+g)(\alpha)=f(\alpha)+g(\alpha)$
(ii) $(cf)(\alpha)=c[f(\alpha)]$
$\forall \ \alpha, \beta \in V, c\in F \ and \ f, g \in L(V, F)$
If $V$ is a finite dimensional vector space over the field $F$, then $dim \ V=dim \ L(V, F)$.

The space all linear functionals i.e. $L(V. F)$ is denoted by $V^*$. Then $dim \ V^* =dim \ V$, if $V$ is finite dimensional vector space. Let $B=\{\alpha_1, \alpha_2, ... , \alpha_n\}$ be an ordered basis for an n-dimensional vector space $V$ over the field $F$. Then we can determine n distinct linear functional $f_1, f_2, ..., f_n$ from $B$ such that
$f_i(\alpha_j)=\delta_{ij}$       where   $\delta_{ij}=0, \text{if} \ i\ne j$
$=1, \text{if} \ i=j$
Claim that: $\{f_1, f_2, ..., f_n\}$ is a linearly independent set in $V^*$.
Consider, $f=\sum c_if_i$
$\implies f(\alpha_j)=\left(\sum c_if_i\right)(\alpha_j)$
$=\sum c_if_i(\alpha_j)$
$=\sum c_i\delta_{ij}$
$=c_j$
Then $f(\alpha_j)=c_j$ ,         for each j.
If $f=0$ then $c_j=0$       for all j.
Hence  $\{f_1, f_2, ..., f_n\}$ is a linearly independent.
Since $dim \ V=n=dim \ V^*$
$\therefore B^*= \{f_1, f_2, ..., f_n\}$ is the basis for $V^*$. This particular basis for $V^*$ is called dual basis.