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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.Â
Let T be a linear transformation for V onto W. Then T is determined by its action on each vector ​\( \alpha_j \in V \)​.Â
Each ​\( T(\alpha_j) \)​ can be expressed asÂ
​\( T(\alpha_j)=\displaystyle\sum_{i=1}^{m}A_{ij}\beta_i   \ \ \ \ (1\le j \le n) \)​Â
where ​\( A_{1j}, A_{2j}, … , A_{mj} \)​ are scalars and are the coordinates of ​\( T(\alpha_j) \)​Â
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’.Â
Conversely, if ​\( \alpha \in V \)​, then
​\( \alpha=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n \)​Â
​\( =\displaystyle\sum_{j=1}^{n}a_j\alpha_j \)​Â
​\( \implies T(\alpha)=T\left(\displaystyle\sum_{j=1}^{n}a_j\alpha_j\right) \)​
            ​\( =\displaystyle\sum_{j=1}^{n}a_jT(\alpha_j) \)​
​\( =\displaystyle\sum_{j=1}^{n}\displaystyle\sum_{i=1}^{m}a_jA_{ij}\beta_i \)​
​\( =\displaystyle\sum_{j=1}^{n}\displaystyle\sum_{i=1}^{m}a_jA_{ij}\beta_i \)​
​\( \implies T(\alpha)=\displaystyle\sum_{j=1}^{n}\displaystyle\sum_{i=1}^{m}a_jA_{ij}\beta_i         \ \ … \ \ (1) \)​
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.Â
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 ​\( F^2 \)​.Â
Solution:Â
Let ​\( B=\{e_1, e_2\} \)​ be the standard basis for the ​\( F^2 \)​ where ​\( e_1=(1, 0), e_2=(0,1) \)​.Â
Then,
​\( T(e_1)=T(1, 0)=(1, 0)=1.e_1+0.e_2 \)​Â
​\( T(e_2)=T(0, 1)=(0, 0)=0.e_1+0.e_2 \)​
Therefore, the matrix of T w.r.t. B is ​\( [T]_B=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \)​
Example:Â
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 ​\( [D]_B \)​.Â
Solution:Â
\[ D(f_1(x))=D(x^0)=D(1)=0=0.f_1+0.f_2+0.f_3+0.f_4 \]
\[ D(f_2(x))=D(x^1)=D(x)=1=1.f_1+0.f_2+0.f_3+0.f_4 \]
\[ D(f_3(x))=D(x^2)=2x=0.f_1+2.f_2+0.f_3+0.f_4 \]
\[ D(f_4(x))=D(x^3)=3x^2=0.f_1+0.f_2+3.f_3+0.f_4 \]
​\( [D]_B=\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix} \)​Â
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 ​\( B=P^{-1}AP \)​.Â
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 ​\( \alpha, \beta \in V \ \text{and} \ c \in F \)​,Â
​\( f(c\alpha+\beta)=cf(\alpha)+f(\beta) \)​
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 ​\( F^n \)​.Â
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,Â
​\( f(cx+y) \)​
​\( =f(c(x_1, x_2, …, x_n)+(y_1, y_2, …, y_n)) \)​
​\( =f((cx_1, cx_2, …, cx_n)+(y_1, y_2, …, y_n)) \)​
​\( =f(cx_1+y_1, cx_2+y_2 …, cx_n+y_n) \)​​
\[ =a_1(cx_1+y_1)+a_2(cx_2+y_2)+ … +a_n(cx_n+y_n)​ \]
\[ =ca_1x_1+a_1y_1+ca_2x_2+a_2y_2+ … +ca_nx_n+a_ny_n \]
\[ =(ca_1x_1+ca_2x_2+…+ca_nx_n)+(a_1y_1+a_2y_2+ … +a_ny_n) \]
\[ =c(a_1x_1+a_2x_2+…+a_nx_n)+(a_1y_1+a_2y_2+ … +a_ny_n) \]
​\( =cf(x_1, x_2, …, x_n)+f(y_1, y_2, …, y_n) \)​
​\( \therefore f(cx+y)=cf(x)+f(y) \)​
​\( \therefore \)​ f is linear functional.Â
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 ​\( trace \ (A)=A_{11}+A_{22}+ … +A_{nn} \)​.Â
Example:Â
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 ​\( dim \ V=dim \ L(V, F) \)​.Â
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 Â
​\( f_i(\alpha_j)=\delta_{ij} \)​   where ​\( \delta_{ij}=0 \)​, ​\( \text{if} \ i\ne j \)​Â
                           ​\( =1, \text{if} \ i=j \)​
Claim that: ​\( \{f_1, f_2, …, f_n\} \)​ is a linearly independent set in\( V^* \).Â
Consider, ​\( f=\sum c_if_i \)​Â
\( \implies f(\alpha_j)=\left(\sum c_if_i\right)(\alpha_j) \)
             \( =\sum c_if_i(\alpha_j) \)
             ​\( =\sum c_i\delta_{ij} \)​
             ​\( =c_j \)​
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.Â
Since ​\( dim \ V=n=dim \ V^* \)​Â
​\( \therefore B^*= \{f_1, f_2, …, f_n\} \)​ is the basis for ​\( V^* \)​. This particular basis for ​\( V^* \)​ is called dual basis.Â
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