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HomeMathematicsAlgebraLinear Transformation Part II - Inverse Linear Transformation and Isomorphism

# Linear Transformation Part II – Inverse Linear Transformation and Isomorphism

#### 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. $Vcong F^n$.
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