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HomeMathematicsAlgebraLinear Transformation Part II - Inverse Linear Transformation and Isomorphism

Linear Transformation Part II – Inverse Linear Transformation and Isomorphism

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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