__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’.