# First isomorphism theorem for vector space, homomorphism, kernel and isomorphism of vector space

## Homomorphism of vector spaces :

Let U & V be the two real vector spaces then a mapping $f:U\to V$ is said to be homomorphism (Linear Transformation) if
(i) $f(\alpha+\beta)=f(\alpha)+f(\beta), \forall \alpha, \beta\in U$
(ii) $f(c\alpha)=cf(\alpha), \forall c\in\mathbb{R}$ & $\alpha\in U$

Combine condition:
$f(a\alpha+b\beta)=af(\alpha)+bf(\beta), \forall a, b\in\mathbb{R}$ & $\alpha, \beta\in U$

## Kernel of Homomorphism :

Let $f:U\to V$ be a homomorphism of real vector space then kernel of f is defined as
$ker f=\{\alpha\in U |f(\alpha)=0\}$, where 0 is the zero vector of vector space V.

## Isomorphism of Vector Space :

Let $V$ be a vector space over $\mathbb{R}$ and $V'$ be another real vector space then a mapping $f:V\to V'$ is called isomorphism of V if
(i) f is one-one.
(ii) f is onto.
(iii) f is homomorphism
and those two vector spaces $V$ & $V'$ are said to be isomorphic and denoted by $V(\mathbb{R})\cong V'(\mathbb{R})$.

## Proof :

Let $T:V\to V'$ be a linear transformation of real vector space $V$ onto real vector space $V'$.
$W=\ker T=\{\alpha\in V | T(\alpha)=0$, where 0 is zero vector in $V'\}$
Here, $W$ is nonempty.   ($\because$ T is linear transportation $\implies T(0)=0'$)
To prove that $W$ is subspace of vector space V, for this, it is enough to prove that,
$\forall a, b\in\mathbb{R}$ and $\forall x, y\in W$, $ax+by\in W$
Let $a, b\in\mathbb{R}$, & $x, y\in W$ be arbitrary.
To prove that $ax+by\in W=\ker T$
Consider, $T(ax+by)=aT(x)+bT(y)$
$=a.0'+b.0'$
$=0'+0'$
$=0'$
$\implies ax+by\in W$
Thus, $\frac{V}{W}$ is quotient space.
Now, to prove that $\frac{V}{W}$ is isomorphism to V'.
For this, consider a mapping $f:\frac{V}{W}\to V'$ such that $f(W+\alpha)=T(\alpha), \forall \alpha\in V$
First to prove that, f is well defined.
Let $W+\alpha, W+\beta$ be any two elements of $\frac{V}{W}$ then $\alpha, \beta\in V$.
Consider, $W+\alpha=W+\beta$
$\implies W+\alpha+\beta=W$
$\implies \alpha+\beta\in W$           ($\because$ W is subspace)
$\implies T(\alpha-\beta)=0'$           (By definition of W=ker T)
$\implies T(\alpha)-T(\beta)=0'$      ($\because$ T is linear)
$\implies T(\alpha)=T(\beta)$
$\implies f(W+\alpha)=f(W+\beta)$     (By definition of f)
$\therefore f:\frac{V}{W}\to V'$ is well defined.
Now, to prove that, $f(W+\alpha)=f(W+\beta)\implies W+\alpha=W+\beta$,     $\forall \ W+\alpha, W+\beta\in \frac{V}{W}$
$f(W+\alpha)=f(W+\beta)\implies T(\alpha)=T(\beta)$
$\implies T(\alpha)-T(\beta)=0'$
$\implies T(\alpha-\beta)=0'$
$\implies \alpha-\beta\in W$
$\implies W+\alpha-\beta=W$      ($\because$ W is subspace)
$\implies W+\alpha=W+\beta$
To prove that, f is onto.
Let x be any vector in vector space $V'$.
As $T:V\to V'$ is onto and $x\in V', \exists \ \alpha$ in vector space $V$ such that $x=T(\alpha)$
$\because \alpha\in V$, we have $(W+\alpha)\in \frac{V}{W}$ such that $f(W+\alpha)=T(\alpha)=x$.
$\therefore \forall \ x\in V', \exists \ W+\alpha\in \frac{V}{W}$ such that $f(W+\alpha)=x$.
Now, to prove that, $f:\frac{V}{W}\to V'$ is linear transformation.
i.e. $f(ax+by)=af(x)+bf(y), a, b\in\mathbb{R}, x, y\in\frac{V}{W}$
As $x, y\in\frac{V}{W}\implies x=W+\alpha, \alpha\in V$ and $y=W+\beta, \beta\in V$
$\therefore f(ax+by)=f[a(W+\alpha)+b(W+\beta)]$
$=f(W+a\alpha+W+b\beta)$
$=f(W+a\alpha+b\beta)$
$=T(a\alpha+b\beta)$
$=aT(\alpha)+bT(\beta)$
$=af(W+\alpha)+bf(W+\beta)$
$=af(x)+bf(y)$
$\therefore$ f is linear.
Hence, f is isomorphism.
$\therefore$ quotient space $\frac{V}{W}$ is isomorphic to $V'$.