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HomeMathematicsAlgebraLinear Transformation Part II - Inverse Linear Transformation and Isomorphism

Linear Transformation Part II – Inverse Linear Transformation and Isomorphism

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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. ​\( V\cong F^n \)​. 
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