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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 $T^{-1}(abeta_1+beta_2)=aT^{-1}(beta_1)+T^{-1}(beta_2)$

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,

$therefore T(aalpha_1+alpha_2)=aT(alpha_1)+T(alpha_2)$

$=abeta_1+beta_2$

$because aalpha_1+alpha_2$ is unique,

$therefore aalpha_1+alpha_2=T^{-1}(abeta_1+beta_2)$

$therefore T^{-1}(abeta_1+beta_2)=aT^{-1}(beta_1)+T^{-1}(beta_2)$

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,

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

$implies T(c_1alpha_1, c_2alpha_2, … , c_nalpha_n)=0$

$implies c_1alpha_1, c_2alpha_2, … , c_nalpha_n=0$ ($because T$ 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 $alpha ne 0in V$

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

__Proof__:

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

Then any $alpha in V$ can be expressed as $alpha=a_1alpha_1+a_2alpha_2+ … +a_nalpha_n$, $forall a_i in F (1le i le n)$.

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 $a in F$

$therefore alpha=a_1alpha_1+a_2alpha_2+ … +a_nalpha_n$, $beta=b_1alpha_1+b_2alpha_2+ … +b_nalpha_n, a_i, b_i in F$

$therefore theta(alpha)=(a_1, a_2, … , a_n)$ and $theta(beta)=(b_1, b_2, … , b_n)$

(i) Consider,

$theta(aalpha+beta)$

$=theta[a(a_1alpha_1+a_2alpha_2+ … +a_nalpha_n)+(b_1alpha_1+b_2alpha_2+ … +b_nalpha_n)]$

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

$=(aa_1+b_1, aa_2+b_2, … ,aa_n+b_n)$

$=a(a_1, a_2, … , a_n)+(b_1, b_2, … , b_n)$

$=atheta(alpha)+theta(beta)$

$therefore theta$ 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 $b_1alpha_1+b_2alpha_2+ … +b_nalpha_n=alpha in V$

$therefore theta(alpha)=theta(b_1alpha_1+b_2alpha_2+ … +b_nalpha_n)$

$=(b_1, b_2, … , b_n)$

$=beta$

$therefore theta$ is onto.

Hence, $theta$ is an isomorphism.

i.e. $Vcong F^n$.