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

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 $V(\mathbb{R})\cong V'(\mathbb{R})$. 

First isomorphism theorem for vector space :

(Fundamental theorem of homomorphism of vector space)

Let $T:V\to V'$ be linear transformation of real vector space V onto real vector space $V$ and $W=ker(T)$ then W is subspace of vector space V and $\frac{V}{W}\cong V'$ 

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, $T(ax+by)=aT(x)+bT(y)$ 
$\implies ax+by\in W$ 
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 $y=W+\beta, \beta\in V$ 
$\therefore f(ax+by)=f[a(W+\alpha)+b(W+\beta)]$ 
$\therefore$ f is linear.
Hence, f is isomorphism. 
$\therefore$ quotient space $\frac{V}{W}$ is isomorphic to $V'$.

Post a Comment

If you have any question, please let me know by comment or email. Also, if you want answers of any mathematical problem, please comment me the question. I will post the answer as early as possible.

Previous Post Next Post