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The Matrix of Linear Transformation and Linear Functionals

The Matrix of Linear Transformation and Linear Functionals

Let $V$ be n-dimensional vector space over the field $F$ and $W$ be an m-dimensional vector space over the same field $F$.

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 
$T(alpha_j)=displaystylesum_{i=1}^{m}A_{ij}beta_i$      $(1le j le n)$ 
where $A_{1j}, A_{2j}, … , A_{mj}$ are scalars and are the coordinates of $T(alpha_j)$ 
The $mtimes n$ matrix A defined by $A(i, j)=A_{ij}, (1le i le m & 1le j le n)$ is the matrix of $T$ relative to the basis $B$ and $B’$. 
Conversely, if $alpha in V$, then
$alpha=a_1alpha_1+a_2alpha_2+ … +a_nalpha_n$ 
$=displaystylesum_{j=1}^{n}a_jalpha_j$ 
$implies T(alpha)=Tleft(displaystylesum_{j=1}^{n}a_jalpha_jright)$ 
                         $=displaystylesum_{j=1}^{n}a_jT(alpha_j)$ 
                         $=displaystylesum_{j=1}^{n}displaystylesum_{i=1}^{m}a_jA_{ij}beta_i$ 
$implies T(alpha)=displaystylesum_{j=1}^{n}displaystylesum_{i=1}^{m}a_jA_{ij}beta_i$                 ……. (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 $alpha, beta in V text{and} c in F$, 
$f(calpha+beta)=cf(alpha)+f(beta)$ 

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 $j (1le jle n)$ 
Now, $f(x_1, x_2, …, x_n)=fleft(displaystylesum_{j=1}^{n}x_je_jright)$ 
                                                $=displaystylesum_{j=1}^{n}x_jf(e_j)$ 
                                                $=displaystylesum_{j=1}^{n}a_jx_j$ 

Definition:

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

Example

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

Solution:

Let $A, B in F^{ntimes n}$ and $cin F$. 
$tr(cA+B)=cA_{11}+cA_{22}+ … +cA_{nn}+B_{11}+B_{22}+ … +B_{nn}$ 
$=displaystylesum_{i=1}^{n}left(cA_{ii}+B_{ii}right)$ 
$=cdisplaystylesum_{i=1}^{n}A_{ii}+displaystylesum_{i=1}^{n}B_{ii}$ 
$=c tr(A)+tr(B)$ 

Note:

If $V$ is a vector space over the field $F$, the set of all linear functionals $f:Vrightarrow F$ 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) $(cf)(alpha)=c[f(alpha)]$ 
$forall alpha, beta in V, cin F and f, g in L(V, F)$ 
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   $delta_{ij}=0,   text{if} ine j$ 
                                                      $=1, text{if} i=j$ 
Claim that: ${f_1, f_2, …, f_n}$ is a linearly independent set in $V^*$. 
Consider, $f=sum c_if_i$ 
$implies f(alpha_j)=left(sum c_if_iright)(alpha_j)$  
                          $=sum c_if_i(alpha_j)$  
                          $=sum c_idelta_{ij}$ 
                          $=c_j$ 
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. 
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