Definition (1):
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 (2):
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 (1):
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 (3):
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 (2):
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
\( T^{-1}(s_1, s_2)=(x_1, x_2) \)
\( T^{-1}(s_1, s_2)=(s_2, s_1-s_2) \)
Also Read: Linear Transformation Part I – Algebra of Linear Transformation
Definition (4):
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 (3):
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. \( V\cong F^n \).
