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# 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
$$T(\alpha_j)=\displaystyle\sum_{i=1}^{m}A_{ij}\beta_i \ \ \ \ (1\le j \le n)$$​
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.
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