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

## 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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