This experiment helps students understand what a linear transformation is and how it is related to matrices. The experiment shows that every linear transformation can be written using a matrix and vice-versa, which makes it easier to work with. By doing this, students learn how matrices can be used to represent and apply transformations in a simple and clear way.

1. Linear transformation:

Let $M$ and $N$ be vector spaces over a field $R$. Then a function $T:M→N$ is called a linear transformation if, for $x, y \in M$ and $\in R$
(i) $T(x+y)=T(x)+T(y)$
(ii) $T(αx)=αT(x)$

1.1. Identity transformation:

Let $M$ be a vector space over a field $R$. Then the function $T:M→M$ defined as $T(x)=x$, where $x \in M$, is linear and is known as identity transformation.

1.2. Zero transformation:

Let $M$ and $N$ be vector spaces over a field $R$. Then the function $T:M→N$ defined as $T(x)=0$, where $x \in M$, is linear and is known as zero transformation.

1.3. Examples:

a). Let $T: R^{2}→R^{2}$ such that $T(x, y)=(x, -y)$, where $x, y \in R$. Then $T$ is linear.

Proof:

Let $x=(x_{1}, x_{2})$ and $y=(y_{1}, y_{2}) \in R^{2}.$ Then $T(x_{1}, x_{2})=(x_{1}, -x_{2})$ and $T(y_{1}, y_{2})=(y_{1}, -y_{2})$. Notice that
(i) $T(x+y)=T[(x_{1}, x_{2}>)+(y_{1}, y_{2})]=T(x_{1}+y_{1}, x_{2}+y_{2})=(x_{1}+y_{1}, -(x_{2}+y_{2}))$ and $T(x)+T(y)=(x_{1}, -x_{2})+ (y_{1}, -y_{2})= (x_{1}+y_{1}, -(x_{2}+y_{2})).$
(ii) $T(αx)=T(αx_{1}, αx_{2})=(αx_{1}, -αx_{2})$ and $αT(x)=α(x_{1}, -x_{2})=(αx_{1}, -αx_{2}).$

b). Let $T:R^{2}$ → $R^{2}$ such that $T(x, y)=(x, 0),$ where $x, y \in R.$ Then $T$ is linear.

Proof:

Let $x=(x_{1}, x_{2})$ and $y=(y_{1}, y_{2}) \in R^{2}$. Then $T(x_{1}, x_{2})=(x_{1}, -x_{2})$ and $T(y_{1}, y_{2})=(y_{1}, -y_{2}).$ Notice that
(i) $T(x+y)=T[(x_{1}, x_{2})+(y_{1}, y_{2})]=T(x_{1}+y_{1}, x_{2}+y_{2})=(x_{1}+y_{1}, 0)$ and $T(x)+T(y)=(x_{1}, 0)+ (y_{1}, 0)=(x_{1}+y_{1}, 0).$
(ii) $T(αx)=T(αx_{1}, αx_{2})=(αx_{1}, 0)$ and $αT(x)=α(x_{1}, 0)=(αx_{1}, 0).$

c). Let $T:R^{2}$ → $R^{2}$ such that $T(x, y)=(x, 0),$ where $x, y \in R.$ Then $T$ is linear.

Proof:

Let $x=(x_{1}, x_{2})$ and $y=(y_{1}, y_{2}) \in R^{2}$. Then $T(x_{1}, x_{2})=(x_{1}, -x_{2})$ and $T(y_{1}, y_{2})=(y_{1}, -y_{2}).$ Notice that
(i) $T(x+y)=T[(x_{1}, x_{2})+(y_{1}, y_{2})]=T(x_{1}+y_{1}, x_{2}+y_{2})=( x_{2}+y_{2}, x_{1}+y_{1})$ and $T(x)+T(y)=(-x_{2}, x_{1})+ (-y_{2}, y_{1})= ( x_{2}+y_{2}, x_{1}+y_{1}).$
(ii) $T(αx)=T(αx_{1}, αx_{2})=(αx_{1}, -αx_{2})$ and $αT(x)=α(x_{2}, x_{1})=(αx_{1}, -αx_{2}).$

1.4. Proposition:

Let $T:M→N$ be a linear transformation. Then
(i) $T(0)=0;$ where 0 on the L.H.S is the zero vector of $M$ and 0 on the R.H.S is the zero vector of $N.$
(ii) $T(-x)=-T(x),$ for all $x \in M$
(iii) $T(x-y)=T(x)-T(y),$ for all $x, y \in M$
(iv) $T(αx+βy)=αT(x)+βT(y),$ for all $x, y \in M$ and $α, β \in R$

1.5. Proposition:

Let $T:V→W$ be a linear map and $V$ and $W$ be finite dimensional vector spaces over $R$ and
$B= \{e_{1}, e_{2}, …, e_{n}\}$ is a basis for $V$ and let $x= r_{1}f_{1}+r_{2} e_{2}+ ... +r_{n}e_{n},$ where $x \in V, r_{1}, r_{2}, …, r_{2} \in R.$ Then $T(e_{1}), T(e_{2}), …, T(e_{n})$ define $T,$ by $T(x)= rT(e_{1})+sT(e_{2})+ … +tT(e_{n}).$

2. Matrix associated with a linear transformation:

Let T:R³→R² be a linear transformation. Let B₁={f₁, f₂, f₃} and B₂={e₁, e₂} be basis of R³ and R² respectively. Then T(f₁)=ae₁+be₂, T(f₂)=ce₁+de₂ and T(f₃)=ge₁+he₂, for some a, b, c, d, g, h∈R.

Let A=

$$\begin{pmatrix}a & c & e \\\ b & d & f\end{pmatrix}$$

Then matrix A is of order 2×3 and is called the matrix representation of T w.r.t. the basis B₁ and B₂.
In a similar way, one can check that a matrix representation of T:Rⁿ→R^m is of order m×n.

2.1. Example:

Let T:R²→R² be a linear map defined as T(x, y)=(x, -y), where x, y∈R. Then find the matrix associated with the transformation w.r.t. the basis B₁={(1, 0), (0,1)}and B₂={(0, -1), (-1, 0)}.
Let e₁=(1, 0), e₂=(0, 1), f₁=(0, -1) and f₂=(-1, 0). Thus
T(1, 0)=(1, 0)=0.(0, -1)+(-1).(-1, 0) T(0, 1)=(0, -1)= 1.(0, -1)+0.(-1, 0) and hence the matrix representation of T w.r.t. the basis B₁ and B₂ is

3. Linear transformation associated with a matrix:

Let A=

$$\begin{pmatrix}a & c & e \\\ b & d & f\end{pmatrix}$$

be a matrix of order 2×3. Then the linear transformation T:R³→R² associated with the matrix w.r.t. the basis B₁=>={f₁, f₂, f₃} and B₂= B₂={e₁, e₂} of R³ and R²respectively can be obtained as follows:
Define T(f₁)=ae₁+be₂, T(f₂)=ce₁+de₂ and T(f₃)=ge₁+he₂, where a, b, c, d, g, h∈R. Then for x∈V, there exist r, s, t∈R such that x=rf₁+sf₂+tf₃. Thus define T(x)= rT(f₁)+sT(f₂)+tT(f₃), because T has to be linear.

3.1. Example:

Consider A= $\begin{pmatrix}1 & -1 & 0 \\\ 0 & 1 & 1\end{pmatrix}$

Then find the associated linear transformation of A w.r.t. the basis B₁={(1, 0, 0), (-1, 1, 0), (0, 1, 1)} and B₂={(-1, 1), (0, 1)} of R³ and R² respectively.
Define T(1, 0, 0)=1(-1, 1)+0(0, 1)=(-1, 1), T(-1, 1, 0)=-1(-1, 1)+1(0, 1)=(1, 0) and T(0, 1, 1)=0(-1, 1)+1(0, 1)=(0, 1). Since (x, y, z)=a(1, 0, 0)+b(-1, 1, 0)+c(0, 1, 1), hence a-b=x, b+c=y and c=z. By solving these equations we get, a=x+y-z, b=y-z and c=z.
Now, define T:R³→R² by T(x, y, z)= aT(1, 0, 0)+bT(-1, 1, 0)+cT(0, 1, 1)=a(-1, 1)+b(1, 0)+c(0, 1)=(-a+b, a+c)=(-x, x+y), where x, y∈R.
The linear transformation T:R³→R² associated with the matrix 2×3 A w.r.t. the basis B₁ and B₂ is T(x, y, z)=( -x, x+y), where x, y∈R.