Linear Transformation

15 MIN READ ADVANCED

Linear Transformation – Definition and Examples

Learning Objectives

• Master derivations of Linear Transformation – Definition and Examples.
• Bridge theoretical limits with practice.

Linear Transformation

Let

V

and

W

be two vector spaces over the same field

F

. A function

T: V \rightarrow W

is said to be a linear transformation if for all

x, y \in V

and

c \in F

, the following conditions are satisfied:

T(x+y)=T(x)+T(y),~\forall~x,y\in V

T(cx)=cT(x),~\forall~x\in V,~c\in F

theorem

Let

V

and

W

be two vector spaces over the same field

F

and let

T: V \rightarrow W

be a linear transformation. Then

T(\overline{\theta}_{V})=\overline{\theta}_{W}.

proof

It is given that

V

and

W

are two vector spaces over the same field

F

and

T: V \rightarrow W

is a linear transformation.

To prove that

T(\overline{\theta}_{V})=\overline{\theta}_{W}

.

We have

\begin{aligned} &\overline{\theta}_{V}+\overline{\theta}_{V} =\overline{\theta}_{V} \\ \implies\;& T(\overline{\theta}_{V}+\overline{\theta}_{V}) =T(\overline{\theta}_{V}) \\ \implies\;& T(\overline{\theta}_{V})+T(\overline{\theta}_{V}) =T(\overline{\theta}_{V}),~~\because~T~\text{is linear} \\ \implies\;& -T(\overline{\theta}_{V})+T(\overline{\theta}_{V}) +T(\overline{\theta}_{V}) =-T(\overline{\theta}_{V})+T(\overline{\theta}_{V}) \\ \implies\;& \overline{\theta}_{W}+T(\overline{\theta}_{V}) =\overline{\theta}_{W} \\ \implies\;& T(\overline{\theta}_{V}) =\overline{\theta}_{W}. \end{aligned}

Hence proved.

\blacksquare

theorem

Let

V

and

W

be two vector spaces over the same field

F

and let

T: V \rightarrow W

be a linear transformation. Then

T(-\overline{x})=-T(\overline{x}), \quad \forall~\overline{x}\in V.

proof

It is given that

V

and

W

are two vector spaces over the same field

F

and

T: V \rightarrow W

is a linear transformation.

To prove that

T(-\overline{x})=-T(\overline{x})

.

Let

\overline{x}\in V

. We have

\begin{aligned} &\overline{x}+(-\overline{x}) =\overline{\theta}_{V} \\ \implies\;& T(\overline{x}+(-\overline{x})) =T(\overline{\theta}_{V}) \\ \implies\;& T(\overline{x})+T(-\overline{x}) =\overline{\theta}_{W},~~\because~T~\text{is linear and } T(\overline{\theta}_{V})=\overline{\theta}_{W} \\ \implies\;& -T(\overline{x})+T(\overline{x}) +T(-\overline{x}) =-T(\overline{x})+\overline{\theta}_{W} \\ \implies\;& T(-\overline{x}) =-T(\overline{x}). \end{aligned}

Hence proved.

\blacksquare

Example-1

Let

T:\mathbb{R}^2 \to \mathbb{R}^2

be defined as

T(a,b)=(0,b),\quad (a,b)\in\mathbb{R}^2.

It is required to determine whether

T

is linear and justify the answer.

Solution

It is given that

T:\mathbb{R}^2 \to \mathbb{R}^2

such that

T(a,b)=(0,b)

.

[1] To prove

T(\overline{x}+\overline{y})=T(\overline{x})+T(\overline{y}),~ \forall~\overline{x},\overline{y}\in\mathbb{R}^2

.

Let

\overline{x},\overline{y}\in\mathbb{R}^2

where

\overline{x}=(a,b)

and

\overline{y}=(c,d)

.

Then

\overline{x}+\overline{y} =(a,b)+(c,d) =(a+c,b+d).

Now

\begin{aligned} T(\overline{x}+\overline{y}) &=T(a+c,b+d)\\ &=(0,b+d)\\ &=(0,b)+(0,d)\\ &=T(a,b)+T(c,d)\\ &=T(\overline{x})+T(\overline{y}). \end{aligned}

[2] To prove

T(k\overline{x})=kT(\overline{x}),~ \forall~\overline{x}\in\mathbb{R}^2,~k\in\mathbb{R}

.

Let

\overline{x}=(a,b)\in\mathbb{R}^2

and

k\in\mathbb{R}

.

Then

k\overline{x}=(ka,kb).

Now

\begin{aligned} T(k\overline{x}) &=T(ka,kb)\\ &=(0,kb)\\ &=k(0,b)\\ &=kT(a,b)\\ &=kT(\overline{x}). \end{aligned}

Hence,

T

is a linear transformation.

\blacksquare

Example-2

Let

T:\mathbb{R}^3 \to \mathbb{R}^2

be defined by

T(a_1,a_2,a_3)=(a_1-a_2,2a_3).

It is required to determine whether

T

is linear.

Solution

It is given that

T:\mathbb{R}^3 \to \mathbb{R}^2

such that

T(a_1,a_2,a_3)=(a_1-a_2,2a_3)

.

[1] To prove additivity.

Let

\overline{x}=(a_1,a_2,a_3)

and

\overline{y}=(b_1,b_2,b_3)

.

Then

\overline{x}+\overline{y} =(a_1+b_1,a_2+b_2,a_3+b_3).

Now

\begin{aligned} T(\overline{x}+\overline{y}) &=((a_1+b_1)-(a_2+b_2),2(a_3+b_3))\\ &=(a_1-a_2,2a_3)+(b_1-b_2,2b_3)\\ &=T(\overline{x})+T(\overline{y}). \end{aligned}

[2] To prove homogeneity.

Let

\overline{x}=(a_1,a_2,a_3)

and

k\in\mathbb{R}

\begin{aligned} T(k\overline{x}) &=T(ka_1,ka_2,ka_3)\\ &=(ka_1-ka_2,2ka_3)\\ &=k(a_1-a_2,2a_3)\\ &=kT(\overline{x}). \end{aligned}

Hence,

T

is a linear transformation.

\blacksquare

Example-3

Let

T:\mathbb{R}^2 \rightarrow \mathbb{R}^3

be defined by

T(a_1,a_2)=(a_1+a_2,0,2a_1-a_2).

It is required to determine whether

T

is linear and justify the answer.

Solution

It is given that

T:\mathbb{R}^2 \to \mathbb{R}^3

such that

T(\overline{a})=(a_1+a_2,0,2a_1-a_2)

, where

\overline{a}=(a_1,a_2)

.

[1] To prove

T(\overline{x}+\overline{y})=T(\overline{x})+T(\overline{y}),~ \forall~\overline{x},\overline{y}\in\mathbb{R}^2

.

Let

\overline{x},\overline{y}\in\mathbb{R}^2

such that

\overline{x}=(a_1,a_2)

and

\overline{y}=(b_1,b_2)

.

Then

\overline{x}+\overline{y} =(a_1+b_1,a_2+b_2).

Now

\begin{aligned} T(\overline{x}+\overline{y}) &=T(a_1+b_1,a_2+b_2)\\ &=\big((a_1+b_1)+(a_2+b_2),0,2(a_1+b_1)-(a_2+b_2)\big)\\ &=\big((a_1+a_2)+(b_1+b_2),0,(2a_1-a_2)+(2b_1-b_2)\big)\\ &=(a_1+a_2,0,2a_1-a_2)+(b_1+b_2,0,2b_1-b_2)\\ &=T(\overline{x})+T(\overline{y}). \end{aligned}

[2] To prove

T(k\overline{x})=kT(\overline{x}),~ \forall~\overline{x}\in\mathbb{R}^2,~k\in\mathbb{R}

.

Let

\overline{x}\in\mathbb{R}^2

where

\overline{x}=(a_1,a_2)

and let

k\in\mathbb{R}

.

Then

k\overline{x} =(ka_1,ka_2).

Now

\begin{aligned} T(k\overline{x}) &=T(ka_1,ka_2)\\ &=(ka_1+ka_2,0,2ka_1-ka_2)\\ &=k(a_1+a_2,0,2a_1-a_2)\\ &=kT(\overline{x}). \end{aligned}

Hence,

T

is a linear transformation.

\blacksquare

Example-4

Let

T: M_{2\times3}(F) \rightarrow M_{2\times2}(F)

be defined by

T(\overline{A})= \begin{pmatrix} a_{11}+a_{12} & a_{13} \\ a_{21} & a_{22}+a_{23} \end{pmatrix},

where

\overline{A}= \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix} \in M_{2\times3}(F).

It is required to determine whether

T

is linear and justify the answer.

Solution

It is given that

T: M_{2\times3}(F)\to M_{2\times2}(F)

such that

T(\overline{A})= \begin{pmatrix} a_{11}+a_{12} & a_{13} \\ a_{21} & a_{22}+a_{23} \end{pmatrix}.

[1] To prove

T(\overline{A}+\overline{B})=T(\overline{A})+T(\overline{B}),~ \forall~\overline{A},\overline{B}\in M_{2\times3}(F)

.

Let

\overline{A}= \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix}, \quad \overline{B}= \begin{pmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \end{pmatrix}.

Then

\overline{A}+\overline{B} = \begin{pmatrix} a_{11}+b_{11} & a_{12}+b_{12} & a_{13}+b_{13} \\ a_{21}+b_{21} & a_{22}+b_{22} & a_{23}+b_{23} \end{pmatrix}.

Now

\begin{aligned} T(\overline{A}+\overline{B}) &= \begin{pmatrix} (a_{11}+b_{11})+(a_{12}+b_{12}) & a_{13}+b_{13} \\ a_{21}+b_{21} & (a_{22}+b_{22})+(a_{23}+b_{23}) \end{pmatrix} \\ &= \begin{pmatrix} (a_{11}+a_{12})+(b_{11}+b_{12}) & a_{13}+b_{13} \\ a_{21}+b_{21} & (a_{22}+a_{23})+(b_{22}+b_{23}) \end{pmatrix} \\ &= \begin{pmatrix} a_{11}+a_{12} & a_{13} \\ a_{21} & a_{22}+a_{23} \end{pmatrix} + \begin{pmatrix} b_{11}+b_{12} & b_{13} \\ b_{21} & b_{22}+b_{23} \end{pmatrix} \\ &= T(\overline{A})+T(\overline{B}). \end{aligned}

[2] To prove

T(k\overline{A})=kT(\overline{A}),~ \forall~\overline{A}\in M_{2\times3}(F),~k\in F

.

Let

k\in F

. Then

k\overline{A} = \begin{pmatrix} ka_{11} & ka_{12} & ka_{13} \\ ka_{21} & ka_{22} & ka_{23} \end{pmatrix}.

Now

\begin{aligned} T(k\overline{A}) &= \begin{pmatrix} ka_{11}+ka_{12} & ka_{13} \\ ka_{21} & ka_{22}+ka_{23} \end{pmatrix} \\ &= k \begin{pmatrix} a_{11}+a_{12} & a_{13} \\ a_{21} & a_{22}+a_{23} \end{pmatrix} \\ &= kT(\overline{A}). \end{aligned}

Hence,

T

is a linear transformation.

\blacksquare

Example-5

Let

T: P_2(\mathbb{R}) \rightarrow P_3(\mathbb{R})

be defined by

T(f(x))=x f(x)+f'(x).

It is required to determine whether

T

is linear and justify the answer.

Solution

It is given that

T: P_2(\mathbb{R}) \to P_3(\mathbb{R})

such that

T(\overline{f})=x\overline{f}(x)+\overline{f}\,'(x),

where

\overline{f}(x)\in P_2(\mathbb{R})

.

[1] To prove

T(\overline{f}+\overline{g})=T(\overline{f})+T(\overline{g}),~ \forall~\overline{f},\overline{g}\in P_2(\mathbb{R})

.

Let

\overline{f},\overline{g}\in P_2(\mathbb{R})

.

Then

(\overline{f}+\overline{g})(x)=\overline{f}(x)+\overline{g}(x).

Now

\begin{aligned} T(\overline{f}+\overline{g}) &=x(\overline{f}(x)+\overline{g}(x))+(\overline{f}+\overline{g})'(x)\\ &=x\overline{f}(x)+x\overline{g}(x)+\overline{f}\,'(x)+\overline{g}\,'(x)\\ &=\big(x\overline{f}(x)+\overline{f}\,'(x)\big) +\big(x\overline{g}(x)+\overline{g}\,'(x)\big)\\ &=T(\overline{f})+T(\overline{g}). \end{aligned}

[2] To prove

T(k\overline{f})=kT(\overline{f}),~ \forall~\overline{f}\in P_2(\mathbb{R}),~k\in\mathbb{R}

.

Let

\overline{f}\in P_2(\mathbb{R})

and

k\in\mathbb{R}

.

Then

(k\overline{f})(x)=k\overline{f}(x).

Now

\begin{aligned} T(k\overline{f}) &=x(k\overline{f}(x))+(k\overline{f})'(x)\\ &=kx\overline{f}(x)+k\overline{f}\,'(x)\\ &=k\big(x\overline{f}(x)+\overline{f}\,'(x)\big)\\ &=kT(\overline{f}). \end{aligned}

Hence,

T

is a linear transformation.

\blacksquare

Example-6

Let

T: M_{n\times n}(F) \rightarrow F

be defined by

T(A)=\operatorname{tr}(A).

It is required to determine whether

T

is linear and justify the answer.

Solution

It is given that

T: M_{n\times n}(F)\to F

such that

T(\overline{A})=\operatorname{tr}(\overline{A}),

where

\overline{A}\in M_{n\times n}(F)

.

[1] To prove

T(\overline{A}+\overline{B})=T(\overline{A})+T(\overline{B}),~ \forall~\overline{A},\overline{B}\in M_{n\times n}(F)

.

Let

\overline{A}=(a_{ij})

and

\overline{B}=(b_{ij})

be in

M_{n\times n}(F)

.

Then

\overline{A}+\overline{B}=(a_{ij}+b_{ij}).

Now

\begin{aligned} T(\overline{A}+\overline{B}) &=\operatorname{tr}(\overline{A}+\overline{B})\\ &=\sum_{i=1}^{n}(a_{ii}+b_{ii})\\ &=\sum_{i=1}^{n}a_{ii}+\sum_{i=1}^{n}b_{ii}\\ &=\operatorname{tr}(\overline{A})+\operatorname{tr}(\overline{B})\\ &=T(\overline{A})+T(\overline{B}). \end{aligned}

[2] To prove

T(k\overline{A})=kT(\overline{A}),~ \forall~\overline{A}\in M_{n\times n}(F),~k\in F

.

Let

k\in F

and

\overline{A}=(a_{ij})\in M_{n\times n}(F)

.

Then

k\overline{A}=(ka_{ij}).

Now

\begin{aligned} T(k\overline{A}) &=\operatorname{tr}(k\overline{A})\\ &=\sum_{i=1}^{n}ka_{ii}\\ &=k\sum_{i=1}^{n}a_{ii}\\ &=k\,\operatorname{tr}(\overline{A})\\ &=kT(\overline{A}). \end{aligned}

Hence,

T

is a linear transformation.

\blacksquare

Example-7

Let

T: M_{2 \times 3}(\mathbb{R}) \rightarrow M_{2 \times 2}(\mathbb{R})

be defined as

\begin{equation} T\left(\overline{A}\right)= \begin{bmatrix} 2a-b & c+2b\\ 0 & 0 \end{bmatrix}, \end{equation}

where

\overline{A}= \begin{bmatrix} a & b & c\\ d & e & f \end{bmatrix} \in M_{2\times3}(\mathbb{R}).

Determine whether

T

is linear and justify the answer. Let

T: M_{2 \times 3}(\mathbb{R}) \rightarrow M_{2 \times 2}(\mathbb{R})

be defined as

\begin{equation} T\left(\overline{A}\right)= \begin{bmatrix} 2a-b & c+2b\\ 0 & 0 \end{bmatrix}, \end{equation}

where

\overline{A}= \begin{bmatrix} a & b & c\\ d & e & f \end{bmatrix} \in M_{2\times3}(\mathbb{R}).

Determine whether

T

is linear and justify the answer.

Solution. It is given that

T: M_{2\times3}(\mathbb{R}) \to M_{2\times2}(\mathbb{R})

such that

T(\overline{A})= \begin{bmatrix} 2a-b & c+2b\\ 0 & 0 \end{bmatrix}.

[1] Additivity: To prove

T(\overline{A}+\overline{B})=T(\overline{A})+T(\overline{B}),\ \forall\,\overline{A},\overline{B}\in M_{2\times3}(\mathbb{R})

Let

\overline{A}= \begin{bmatrix} a & b & c\\ d & e & f \end{bmatrix}, \qquad \overline{B}= \begin{bmatrix} a_1 & b_1 & c_1\\ d_1 & e_1 & f_1 \end{bmatrix}.

Then

\overline{A}+\overline{B}= \begin{bmatrix} a+a_1 & b+b_1 & c+c_1\\ d+d_1 & e+e_1 & f+f_1 \end{bmatrix}.

Now,

\begin{aligned} T(\overline{A}+\overline{B}) &= \begin{bmatrix} 2(a+a_1)-(b+b_1) & (c+c_1)+2(b+b_1)\\ 0 & 0 \end{bmatrix}\\ &= \begin{bmatrix} (2a-b)+(2a_1-b_1) & (c+2b)+(c_1+2b_1)\\ 0 & 0 \end{bmatrix}\\ &= \begin{bmatrix} 2a-b & c+2b\\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 2a_1-b_1 & c_1+2b_1\\ 0 & 0 \end{bmatrix}\\ &= T(\overline{A})+T(\overline{B}). \end{aligned}