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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, ​\( T(ax+by)=aT(x)+bT(y) \)​ 
                                              ​\( =a.0’+b.0′ \)
                                              ​\( =0’+0′ \)
                                              ​\( =0′ \)
\( \implies ax+by\in W \)
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’.
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