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HomeMathematicsAlgebraThe 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 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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