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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 â€‹\( B=\{\alpha_1, \alpha_2, … , \alpha_n\} \)â€‹ be an ordered basis for V and â€‹\( B’=\{\beta_1, \beta_2, … , \beta_n\} \)â€‹ 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 â€‹\( \alpha_j \in V \)â€‹.Â

Each â€‹\( T(\alpha_j) \)â€‹ can be expressed asÂ

â€‹\( T(\alpha_j)=\displaystyle\sum_{i=1}^{m}A_{ij}\beta_iÂ Â Â \ \ \ \ (1\le j \le n) \)â€‹Â

where â€‹\( A_{1j}, A_{2j}, … , A_{mj} \)â€‹ are scalars and are the coordinates of â€‹\( T(\alpha_j) \)â€‹Â

The â€‹\( m\times n \)â€‹ matrix A defined by â€‹\( A(i, j)=A_{ij}, (1\le i \le m \ \& \ 1\le j \le n) \)â€‹ is the matrix of T relative to the basis B and B’.Â

Conversely, if â€‹\( \alpha \in V \)â€‹, then

â€‹\( \alpha=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n \)â€‹Â

â€‹\( =\displaystyle\sum_{j=1}^{n}a_j\alpha_j \)â€‹Â

â€‹\( \implies T(\alpha)=T\left(\displaystyle\sum_{j=1}^{n}a_j\alpha_j\right) \)â€‹

Â Â Â Â Â Â Â Â Â Â Â Â â€‹\( =\displaystyle\sum_{j=1}^{n}a_jT(\alpha_j) \)â€‹

â€‹\( =\displaystyle\sum_{j=1}^{n}\displaystyle\sum_{i=1}^{m}a_jA_{ij}\beta_i \)â€‹

â€‹\( \implies T(\alpha)=\displaystyle\sum_{j=1}^{n}\displaystyle\sum_{i=1}^{m}a_jA_{ij}\beta_iÂ Â Â Â Â Â Â Â Â \ \ … \ \ (1) \)â€‹

If A is â€‹\( m \times n \)â€‹ 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 â€‹\( F^2 \)â€‹ as â€‹\( T(x_1, x_2)=(x_1, 0) \)â€‹. Then find the matrix T w.r.t. the standard ordered basis for â€‹\( F^2 \)â€‹.Â

**Solution**:Â

Let â€‹\( B=\{e_1, e_2\} \)â€‹ be the standard basis for the â€‹\( F^2 \)â€‹ where â€‹\( e_1=(1, 0), e_2=(0,1) \)â€‹.Â

Then,

â€‹\( T(e_1)=T(1, 0)=(1, 0)=1.e_1+0.e_2 \)â€‹Â

â€‹\( T(e_2)=T(0, 1)=(0, 0)=0.e_1+0.e_2 \)â€‹

Therefore, the matrix of T w.r.t. B is â€‹\( [T]_B=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \)â€‹

__Example__:Â

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

__Solution__:Â

\[ D(f_1(x))=D(x^0)=D(1)=0=0.f_1+0.f_2+0.f_3+0.f_4 \]

\[ D(f_2(x))=D(x^1)=D(x)=1=1.f_1+0.f_2+0.f_3+0.f_4 \]

\[ D(f_3(x))=D(x^2)=2x=0.f_1+2.f_2+0.f_3+0.f_4 \]

\[ D(f_4(x))=D(x^3)=3x^2=0.f_1+0.f_2+3.f_3+0.f_4 \]

â€‹\( [D]_B=\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\Â 0 & 0 & 0 & 0 \end{bmatrix} \)â€‹Â

__Definition__:

Let A and B be â€‹\( m \times n \)â€‹ matrices over the field F. Then B is said to be similar to A if there exists an invertible matrix P such that â€‹\( B=P^{-1}AP \)â€‹.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. â€‹\( f:V \rightarrow F \)â€‹ is called a linear functional if for any vector â€‹\( \alpha, \beta \in V \ \text{and} \ c \in F \)â€‹,Â

â€‹\( f(c\alpha+\beta)=cf(\alpha)+f(\beta) \)â€‹

__Example__:

Let F be a field and for some scalars â€‹\( a_1, a_2, … , a_n \)â€‹ define f on â€‹\( F^n \)â€‹ by â€‹

\[ f(x_1, x_2, … , x_n)=a_1x_1+a_2x_2+ … +a_nx_n. \]

Then show that f is a linear functional on â€‹\( F^n \)â€‹.Â

__Solution__:

Let\( x, y \in F^n \)

â€‹\( \therefore x=(x_1, x_2, …, x_n)\Â \& \ y=(y_1, y_2, …, y_n) \)â€‹

Let â€‹\( c \in F \)â€‹

Consider,Â

â€‹\( f(cx+y) \)â€‹

â€‹\( =f(c(x_1, x_2, …, x_n)+(y_1, y_2, …, y_n)) \)â€‹

â€‹\( =f((cx_1, cx_2, …, cx_n)+(y_1, y_2, …, y_n)) \)â€‹

â€‹\( =f(cx_1+y_1, cx_2+y_2 …, cx_n+y_n) \)â€‹â€‹

\[ =a_1(cx_1+y_1)+a_2(cx_2+y_2)+ … +a_n(cx_n+y_n)â€‹ \]

\[ =ca_1x_1+a_1y_1+ca_2x_2+a_2y_2+ … +ca_nx_n+a_ny_n \]

\[ =(ca_1x_1+ca_2x_2+…+ca_nx_n)+(a_1y_1+a_2y_2+ … +a_ny_n) \]

\[ =c(a_1x_1+a_2x_2+…+a_nx_n)+(a_1y_1+a_2y_2+ … +a_ny_n) \]

â€‹\( =cf(x_1, x_2, …, x_n)+f(y_1, y_2, …, y_n) \)â€‹

â€‹\( \therefore f(cx+y)=cf(x)+f(y) \)â€‹

â€‹\( \therefore \)â€‹Â f is linear functional.Â

__Note__:

The matrix of f w.r.t. the standard basis â€‹\( B=\{e_1, e_2, … , e_n\} \)â€‹ (where â€‹\( e_i=(0, 0, … , 1_{i^{th}}, … , 0 \)â€‹) and â€‹\( B’=\{1\} \)â€‹ is given by â€‹\( [a_1, a_2, … , a_n] \)â€‹.

Here â€‹\( f(e_j)=a_j \)â€‹, for each â€‹\( j \ (1\le j\le n) \)â€‹

Now, â€‹\( f(x_1, x_2, …, x_n)=f\left(\displaystyle\sum_{j=1}^{n}x_je_j\right) \)â€‹

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â â€‹\( =\displaystyle\sum_{j=1}^{n}x_jf(e_j) \)â€‹

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â â€‹\( =\displaystyle\sum_{j=1}^{n}a_jx_j \)â€‹

__Definition__:

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

__Example__:Â

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

__Solution__:

Let â€‹\( A, B \in F^{n\times n} \)â€‹ and â€‹\( c\in F \)â€‹.Â â€‹

\[ tr(cA+B)=cA_{11}+cA_{22}+ … +cA_{nn}+B_{11}+B_{22}+ … +B_{nn} \]

â€‹â€‹\( =\displaystyle\sum_{i=1}^{n}\left(cA_{ii}+B_{ii}\right) \)â€‹

â€‹\( =c\displaystyle\sum_{i=1}^{n}A_{ii}+\displaystyle\sum_{i=1}^{n}B_{ii} \)â€‹

â€‹\( =c \ tr(A)+tr(B) \)â€‹

__Note__:

If V is a vector space over the field F, the set of all linear functionals â€‹\( f:V\rightarrow F \)â€‹ is denoted by â€‹\( L(V, F) \)â€‹ and â€‹\( L(V, F) \)â€‹ is a vector space over the field F. w.r.t. following

(i) â€‹\( (f+g)(\alpha)=f(\alpha)+g(\alpha) \)â€‹

(ii) â€‹\( (cf)(\alpha)=c[f(\alpha)] \)â€‹

â€‹\( \forall \ \alpha, \beta \in V, c\in F \ and \ f, g \in L(V, F) \)â€‹

If V is a finite dimensional vector space over the field F, then â€‹\( dim \ V=dim \ L(V, F) \)â€‹.Â

The space all linear functionals i.e. â€‹\( L(V, F) \)â€‹ is denoted by â€‹\( V^* \)â€‹. Then â€‹\( dim \ V^* =dim \ V \)â€‹, if V is finite dimensional vector space. Let â€‹\( B=\{\alpha_1, \alpha_2, … , \alpha_n\} \)â€‹ be an ordered basis for an n-dimensional vector space V over the field F. Then we can determine n distinct linear functional â€‹\( f_1, f_2, …, f_n \)â€‹ from B such thatÂ Â

â€‹\( f_i(\alpha_j)=\delta_{ij} \)â€‹Â Â Â whereÂ â€‹\( \delta_{ij}=0 \)â€‹,Â â€‹\( \text{if} \ i\ne j \)â€‹Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â â€‹\( =1, \text{if} \ i=j \)â€‹

**Claim that:** â€‹\( \{f_1, f_2, …, f_n\} \)â€‹ is a linearly independent set in\( V^* \).Â

Consider, â€‹\( f=\sum c_if_i \)â€‹Â

\( \implies f(\alpha_j)=\left(\sum c_if_i\right)(\alpha_j) \)

Â Â Â Â Â Â Â Â Â Â Â Â Â \( =\sum c_if_i(\alpha_j) \)

Â Â Â Â Â Â Â Â Â Â Â Â Â â€‹\( =\sum c_i\delta_{ij} \)â€‹

Â Â Â Â Â Â Â Â Â Â Â Â Â â€‹\( =c_j \)â€‹

Then â€‹\( f(\alpha_j)=c_j \)â€‹ ,Â Â Â Â Â for each j.

If â€‹\( f=0 \)â€‹ then â€‹\( c_j=0 \)â€‹ Â Â Â for all j.

HenceÂ â€‹\( \{f_1, f_2, …, f_n\} \)â€‹Â is a linearly independent.Â

Since â€‹\( dim \ V=n=dim \ V^* \)â€‹Â

â€‹\( \therefore B^*= \{f_1, f_2, …, f_n\} \)â€‹ is the basis for â€‹\( V^* \)â€‹. This particular basis for â€‹\( V^* \)â€‹ is called dual basis.Â

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