27.9 C
Maharashtra
Friday, June 9, 2023
HomeMathematicsAlgebraLinear Transformation Part II - Inverse Linear Transformation and Isomorphism

# Linear Transformation Part II – Inverse Linear Transformation and Isomorphism

## Definition:

A function T from V into W is called invertible if there exists a function S from W to V such that TS is an identity on W and ST is an identity on V.
i.e. ​$$TS=I_w, ST=I_v$$​

## Definition:

T is invertible, if
(i) T is one-one i.e. ​$$T(\alpha)=T(\beta) \implies \alpha=\beta$$
(ii) T is onto i.e. for any ​$$\beta \in W, \exists \ \alpha \in V, \ such \ that \ T(\alpha)=\beta$$

#### Notation:

If T is invertible and S is an inverse of T, then ​$$S=T^{-1}$$​.

## Theorem:

Let V and W be vector spaces over the filed F and T be a linear transformation. Then ​$$T^{-1}$$​ is a linear transformation from W into V.

## Proof:

Claim that ​$$T^{-1}:W \rightarrow V$$​ is a linear transformation.
Let ​$$\beta_1, \beta_2 \in W$$​ be any vectors and ​$$a \in F$$​ be any scalar.
To show that ​$$T^{-1}(a\beta_1+\beta_2)=aT^{-1}(\beta_1)+T^{-1}(\beta_2)$$​
Let ​$$\alpha_i=T^{-1}(\beta_i)$$​
i.e. ​$$\alpha_i$$​ is unique such that ​$$T(\alpha_i)=\beta_i$$​,        ​$$(i=1, 2)$$​
Since, T is linear,
$$\therefore T(a\alpha_1+\alpha_2)=aT(\alpha_1)+T(\alpha_2)$$
​$$=a\beta_1+\beta_2$$
$$\because a\alpha_1+\alpha_2$$​ is unique,
$$\therefore a\alpha_1+\alpha_2=T^{-1}(a\beta_1+\beta_2)$$​
$$\therefore T^{-1}(a\beta_1+\beta_2)=aT^{-1}(\beta_1)+T^{-1}(\beta_2)$$​
Hence, ​$$T^{-1}$$​ is a linear transformation.

#### Definition:

A linear transformation T from V into W is said to be non-singular if$$T(\alpha)=0$$, ​$$\implies \alpha=0, \forall \alpha \in V$$​.
i.e. the null space of T is zero if T is non-singular which concludes that T is one-one iff T is non-singular.

## Theorem:

Let T be a linear transformation from vector space V into vector space W over the field F. T is non-singular iff T carries each linearly independent subset of V onto a linearly independent subset of W.

## Proof:

Suppose that T is non-singular.
Let ​$$S=\{\alpha_1, \alpha_2, … , \alpha_n\}$$​ be a linearly independent subset of a vector space V.
Claim that$$S’=\{T(\alpha_1), T(\alpha_2), … , T(\alpha_n)\}$$ is linearly independent.
Suppose that,
$$c_1T(\alpha_1)+c_2T(\alpha_2)+ … +c_nT(\alpha_n)=0$$,          ​$$c_1, c_2, … , c_n \in F$$
$$\implies T(c_1\alpha_1, c_2\alpha_2, … , c_n\alpha_n)=0$$
$$\implies c_1\alpha_1, c_2\alpha_2, … , c_n\alpha_n=0$$​   (​$$\because$$​ T is  non-singular)
$$\implies c_1=0, c_2=0, … ,c_n=0$$​          (​$$\because$$​ S is linearly independent)

Thus, ​

$c_1T(\alpha_1)+c_2T(\alpha_2)+ … +c_nT(\alpha_n)=0, \implies c_1=0, c_2=0, … ,c_n=0$

Therefore, ​$$S’=\{T(\alpha_1), T(\alpha_2), … , T(\alpha_n)\}$$​ is linearly independent.
Hence, T carries each linearly independent subset of V onto a linearly independent subset of W.
Conversely,
Assume that T carries each linearly independent subset of V onto a linearly independent subset of W.
Claim that T is non-singular.
Suppose that ​$$\alpha \ne 0\in V$$​
Then ​$$\{\alpha\}\subset V$$​ is linearly independent.
$$\implies \{T(\alpha)\}\subset W$$​ is linearly independent.
$$\implies T(\alpha)\ne 0$$​
Thus, ​$$\alpha \ne 0 \implies T(\alpha)\ne 0$$
Hence, T is non-singular.

#### Example:

Let F be a real field and T be a linear transformation defined on ​$$F^2$$​ given by ​$$T(x_1, x_2)=(x_1+x_2, x_1)$$​. Verify that T is non-singular.

#### Solution:

$$T(x_1, x_2)=0$$
$$\implies (x_1+x_2, x_1)=0$$
$$\implies x_1+x_2=0, x_1=0$$​
$$\implies x_1=0, x_2=0$$​
$$\implies (x_1, x_2)=0$$​
$$\therefore T(x_1, x_2)=0 \implies (x_1, x_2)=0$$​
T is non-singular.
Now,
$$T(x_1, x_2)=(x_1+x_2, x_1)=(s_1, s_2)$$​
$$\therefore x_1+x_2=s_1, x_1=s_2$$
$$\therefore x_1=s_2, x_2=s_1-s_2$$​
If ​$$T^{-1}$$​ exists, then
$$T^{-1}(s_1, s_2)=(x_1, x_2)$$
$$T^{-1}(s_1, s_2)=(s_2, s_1-s_2)$$

#### Definition:

Let V and W be vector spaces over the filed F. A linear transformation from V onto W is said to be an isomorphism if
(i) T is one-one
(ii) T is onto
If T is isomorphism then in this case, V is isomorphic to W.

## Theorem:

Let V be an n-dimensional vector space over the filed F. Then there is an isomorphism for V onto ​

$F^n (F^n=\{(x_1, x_2, … , x_n) | x_i\in F \ \text{is a vector space})$

## Proof:

Let$$B=\{\alpha_1, \alpha_2, …, \alpha_n\}$$ be an ordered basis for V.

Then any ​$$\alpha \in V$$​ can be expressed as ​

$\alpha=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n, \forall a_i \in F \ (1\le i \le n).$

Define ​$$\theta: V \rightarrow F^n$$​ by ​$$\theta(\alpha)=(a_1, a_2, … , a_n)$$
Claim: ​$$\theta$$​ is an isomorphism.
Let ​$$\alpha, \beta \in V$$​ and ​$$a \in F$$​
$$\therefore \alpha=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n,$$​ ​$$\beta=b_1\alpha_1+b_2\alpha_2+ … +b_n\alpha_n, \ a_i, b_i \in F$$​
$$\therefore \theta(\alpha)=(a_1, a_2, … , a_n)$$​ and ​$$\theta(\beta)=(b_1, b_2, … , b_n)$$
(i) Consider,
$$\theta(a\alpha+\beta)$$​​

$=\theta[a(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)+(b_1\alpha_1+b_2\alpha_2+ … +b_n\alpha_n)]​$

$=\theta[(aa_1+b_1)\alpha_1+(aa_2+b_2)\alpha_2+ … +(aa_n+b_n)\alpha_n]$

$$=(aa_1+b_1, aa_2+b_2, … ,aa_n+b_n)$$
$$=a(a_1, a_2, … , a_n)+(b_1, b_2, … , b_n)$$
$$=a\theta(\alpha)+\theta(\beta)$$
$$\therefore \theta$$​ is linear transformation.
(ii) Suppose that, ​$$\theta(\alpha)=\theta(\beta)$$​
$$\therefore (a_1, a_2, … , a_n)=(b_1, b_2, … , b_n)$$
$$\therefore a_1=b_1, a_2=b_2, … , a_n=b_n$$
$$\therefore \alpha=\beta$$​.
Hence, ​$$\theta$$​ is one-one.
(iii) Let ​$$\beta \in F^n$$​.
$$\therefore \beta=(b_1, b_2, … , b_n)$$
Take ​$$b_1\alpha_1+b_2\alpha_2+ … +b_n\alpha_n=\alpha \in V$$​
$$\therefore \theta(\alpha)=\theta(b_1\alpha_1+b_2\alpha_2+ … +b_n\alpha_n)$$
​$$=(b_1, b_2, … , b_n)$$
​$$=\beta$$
$$\therefore \theta$$​ is onto.
Hence, \theta is an isomorphism.
i.e. ​$$V\cong F^n$$​.
RELATED ARTICLES

### Any two closed subset of metric space are connected iff they are disjoint

$${}$$