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__Definition__:Â

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.Â

i.e. â€‹\( TS=I_w, ST=I_v \)â€‹Â

__Definition__**:**Â

T is invertible, ifÂ

(i) T is one-one i.e. â€‹\( T(\alpha)=T(\beta) \implies \alpha=\beta \)â€‹

(ii) T is onto i.e. for any â€‹\( \beta \in W, \exists \ \alpha \in V, \ such \ that \ T(\alpha)=\beta \)â€‹

__Notation__:

If T is invertible and S is an inverse of T, then â€‹\( S=T^{-1} \)â€‹.Â

__Theorem__:

**Let V and W be vector spaces over the filed F and T be a linear transformation. Then â€‹\( T^{-1} \)â€‹Â is a linear transformation from W into V.Â **

__Proof__:

Claim that â€‹\( T^{-1}:W \rightarrow V \)â€‹Â is a linear transformation.Â

Â Let â€‹\( \beta_1, \beta_2 \in W \)â€‹ be any vectors and â€‹\( a \in F \)â€‹Â be any scalar.

To show that â€‹\( T^{-1}(a\beta_1+\beta_2)=aT^{-1}(\beta_1)+T^{-1}(\beta_2) \)â€‹Â

Let â€‹\( \alpha_i=T^{-1}(\beta_i) \)â€‹Â

i.e. â€‹\( \alpha_i \)â€‹ is unique such that â€‹\( T(\alpha_i)=\beta_i \)â€‹,Â Â Â Â â€‹\( (i=1, 2) \)â€‹Â

Since, T is linear,

\( \therefore T(a\alpha_1+\alpha_2)=aT(\alpha_1)+T(\alpha_2) \)Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â â€‹\( =a\beta_1+\beta_2 \)â€‹

â€‹\( \because a\alpha_1+\alpha_2 \)â€‹Â is unique,

â€‹\( \therefore a\alpha_1+\alpha_2=T^{-1}(a\beta_1+\beta_2) \)â€‹Â

â€‹\( \therefore T^{-1}(a\beta_1+\beta_2)=aT^{-1}(\beta_1)+T^{-1}(\beta_2) \)â€‹Â

Â Hence, â€‹\( T^{-1} \)â€‹Â is a linear transformation.Â

__Definition__:Â Â

A linear transformation T from V into W is said to be non-singular if\( T(\alpha)=0 \), â€‹\( \implies \alpha=0, \forall \alpha \in V \)â€‹.

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.**Â

__Proof__:Â

Suppose that T is non-singular.Â

Let â€‹\( S=\{\alpha_1, \alpha_2, … , \alpha_n\} \)â€‹Â be a linearly independent subset of a vector space V.Â

Claim that\( S’=\{T(\alpha_1), T(\alpha_2), … , T(\alpha_n)\} \)Â is linearly independent.Â

Suppose that,

\( c_1T(\alpha_1)+c_2T(\alpha_2)+ … +c_nT(\alpha_n)=0 \),Â Â Â Â Â â€‹\( c_1, c_2, … , c_n \in F \)â€‹

â€‹\( \implies T(c_1\alpha_1, c_2\alpha_2, … , c_n\alpha_n)=0 \)â€‹

â€‹\( \implies c_1\alpha_1, c_2\alpha_2, … , c_n\alpha_n=0 \)â€‹Â Â (â€‹\( \because \)â€‹Â T isÂ non-singular)

â€‹\( \implies c_1=0, c_2=0, … ,c_n=0 \)â€‹Â Â Â Â Â (â€‹\( \because \)â€‹Â S is linearly independent)Â

Thus, â€‹

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

Therefore, â€‹\( S’=\{T(\alpha_1), T(\alpha_2), … , T(\alpha_n)\} \)â€‹Â is linearly independent.

Hence, T carries each linearly independent subset of V onto a linearly independent subset of W.Â

Conversely,Â

Assume that T carries each linearly independent subset of V onto a linearly independent subset of W.Â

Claim that T is non-singular.

Suppose that â€‹\( \alpha \ne 0\in V \)â€‹Â

Then â€‹\( \{\alpha\}\subset V \)â€‹Â is linearly independent.Â

â€‹\( \implies \{T(\alpha)\}\subset W \)â€‹Â is linearly independent.Â

â€‹\( \implies T(\alpha)\ne 0 \)â€‹Â

Thus, â€‹\( \alpha \ne 0 \implies T(\alpha)\ne 0 \)â€‹

Hence, T is non-singular.

__Example__:Â

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

__Solution__:

â€‹\( T(x_1, x_2)=0 \)â€‹

â€‹\( \implies (x_1+x_2, x_1)=0 \)â€‹

â€‹\( \implies x_1+x_2=0, x_1=0 \)â€‹Â

â€‹\( \implies x_1=0, x_2=0 \)â€‹Â

â€‹\( \implies (x_1, x_2)=0 \)â€‹Â

â€‹\( \therefore T(x_1, x_2)=0 \implies (x_1, x_2)=0 \)â€‹Â

T is non-singular.Â

Now,Â

â€‹\( T(x_1, x_2)=(x_1+x_2, x_1)=(s_1, s_2) \)â€‹Â

â€‹\( \therefore x_1+x_2=s_1, x_1=s_2 \)â€‹

â€‹\( \therefore x_1=s_2, x_2=s_1-s_2 \)â€‹Â

If â€‹\( T^{-1} \)â€‹Â exists, then

â€‹\( T^{-1}(s_1, s_2)=(x_1, x_2) \)â€‹

â€‹\( T^{-1}(s_1, s_2)=(s_2, s_1-s_2) \)â€‹

Â

__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 ontoÂ

Â Â Â Â Â 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 â€‹

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

__Proof__:Â

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

Then any â€‹\( \alpha \in V \)â€‹ can be expressed as â€‹

\[ \alpha=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n, \forall a_i \in FÂ Â Â \ (1\le i \le n). \]

Define â€‹\( \theta: V \rightarrow F^n \)â€‹ by â€‹\( \theta(\alpha)=(a_1, a_2, … , a_n) \)â€‹

Claim: â€‹\( \theta \)â€‹Â is an isomorphism.

Let â€‹\( \alpha, \beta \in V \)â€‹ and â€‹\( a \in F \)â€‹Â

â€‹\( \therefore \alpha=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n, \)â€‹ â€‹\( \beta=b_1\alpha_1+b_2\alpha_2+ … +b_n\alpha_n, \ a_i, b_i \in F \)â€‹Â

â€‹\( \therefore \theta(\alpha)=(a_1, a_2, … , a_n) \)â€‹ and â€‹\( \theta(\beta)=(b_1, b_2, … , b_n) \)â€‹

(i) Consider,

â€‹\( \theta(a\alpha+\beta) \)â€‹â€‹

\[ =\theta[a(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)+(b_1\alpha_1+b_2\alpha_2+ … +b_n\alpha_n)]â€‹ \]

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

â€‹\( =(aa_1+b_1, aa_2+b_2, … ,aa_n+b_n) \)â€‹

â€‹\( =a(a_1, a_2, … , a_n)+(b_1, b_2, … , b_n) \)â€‹

â€‹\( =a\theta(\alpha)+\theta(\beta) \)â€‹

â€‹\( \therefore \theta \)â€‹Â is linear transformation.

(ii) Suppose that, â€‹\( \theta(\alpha)=\theta(\beta) \)â€‹Â

â€‹\( \therefore (a_1, a_2, … , a_n)=(b_1, b_2, … , b_n) \)â€‹

â€‹\( \therefore a_1=b_1, a_2=b_2, … , a_n=b_n \)â€‹

â€‹\( \therefore \alpha=\beta \)â€‹.

Hence, â€‹\( \theta \)â€‹ is one-one.

(iii) Let â€‹\( \beta \in F^n \)â€‹.

â€‹\( \therefore \beta=(b_1, b_2, … , b_n) \)â€‹

Take â€‹\( b_1\alpha_1+b_2\alpha_2+ … +b_n\alpha_n=\alpha \in V \)â€‹Â

â€‹\( \therefore \theta(\alpha)=\theta(b_1\alpha_1+b_2\alpha_2+ … +b_n\alpha_n) \)â€‹

Â Â Â Â Â Â Â Â â€‹\( =(b_1, b_2, … , b_n) \)â€‹

Â Â Â Â Â Â Â Â â€‹\( =\beta \)â€‹

â€‹\( \therefore \theta \)â€‹Â is onto.

Hence, \theta is an isomorphism.

i.e. â€‹\( V\cong F^n \)â€‹.Â

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