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__Theorem__:Â

**Let V and W be two vector spaces on the same field. Let S and T be linear transformations from V into W. Then â€‹\( S+T \)â€‹ is a linear transformation from V into W w.r.t. â€‹\( (S+T)(\alpha)=S(\alpha)+T(\alpha) , \forall \ \alpha \in V \)â€‹. If c is any scalar in F, then cT is a linear transformation from V into W w.r.t.â€‹\( (cT)(\alpha)=c(T(\alpha)), \forall \ \alpha \in V \)â€‹. The set of all linear transformation from the vector space V into W is a vector space over the field F.**

__Proof__:

Let â€‹\( \alpha, \beta \in V \)â€‹ and c \in F and S & T are linear transformation from V to W.Â

â€‹\( (S+T)(c\alpha+\beta) \)

\( =S(c\alpha+\beta)+T(c\alpha+\beta) \)

\( =cS(\alpha)+S(\beta)+cT(\alpha)+T(\beta) \)

â€‹\( =c(S+T)(\alpha)+(S+T)(\alpha) \)â€‹Â

Â which shows that â€‹\( (S+T) \)â€‹ is a linear transformation.Â

Similarly, â€‹

\[ (cT)(d\alpha+\beta)=c[T(d\alpha+\beta)]=c[dT(\alpha)+T(\beta)] \]

â€‹\( =cd(T(\alpha))+cT(\beta) \)â€‹

â€‹\( =d[cT(\alpha)]+cT(\alpha) \)â€‹

â€‹\( =d[(cT)(\alpha)]+(cT)(\beta) \)â€‹

which shows that\( (cT) \)is a linear transformation.

__Note__:

The set of all linear transformation from a vector space V into W is a vector space over the field F and it is denoted by â€‹\( L(V, W) \)â€‹. Thus, â€‹\( L(V, W)=\{T:V\rightarrow W| T \ is \ a \ L.T.\} \)â€‹ is a vector space over the field F w.r.t.Â

â€‹\( (S+T)(\alpha)=S(\alpha)+T(\alpha) \)â€‹.

â€‹\( (cT)(\alpha)=c(T(\alpha)) \)â€‹

â€‹\( \forall S, T \in L(V, W), \alpha \in V \)â€‹ and â€‹\( c \in F \)â€‹.

Â

__Theorem__:

Let V be an n-dimensional vector space and W be an m-dimensional vector space over the field F then â€‹\( L(V, W) \)â€‹ is finite dimensional and has dimension ‘mn’.

**Proof**:Â

Let â€‹\( B=\{\alpha_1, \alpha_2, … ,\alpha_n\} \)â€‹ and â€‹\( B’=\{\beta_1, \beta_2, … ,\beta_m\} \)â€‹ be ordered basis for V and W respectively.

Then by theorem, there exists a unique linear transformation â€‹\( T_{11} \)â€‹ such that

\[ T_{11}(\alpha_1)=\beta_1, T_{11}(\alpha_2)=0, T_{11}(\alpha_3)=0, … , T_{11}(\alpha_n)=0 \]

where â€‹\( \beta_1, 0, 0, … , 0 \)â€‹ are the vectors in W.

But for any pair â€‹\( (p, q) \)â€‹ where â€‹\( 1\le p \le m \)â€‹ and â€‹\( 1 \le q \le n \)â€‹ there exists a linear transformation â€‹\( T_{pq} \)â€‹ from V into W such that â€‹\( T_{pq} = \begin{cases} 0 & {if \ \ i \ne q} \\ \beta_p & {if \ \ i = q} \end{cases} \)â€‹

i.e. â€‹\( T_{pq}(\alpha_i)=\delta_{iq}\beta_p, \text{where} \ \delta_{iq} = \begin{cases} 1 & {if \ \ i=q} \\ 0 & {if \ \ i \ne q} \end{cases} \)â€‹

Since, â€‹\( 1\le p \le m \)â€‹ and â€‹\( 1 \le q \le n \)â€‹, there are mn such\( T_{pq} \)‘s.Â

Let â€‹

\[ B_1=\{T_{pq}|1\le p \le m \ and \ 1 \le q \le n\} (\# B_1=mn) \]

Claim that â€‹\( B_1 \)â€‹ is the basis for â€‹\( L(V, W) \)â€‹

i.e. to show that (i) â€‹\( L(B_1)=L(V, W) \)â€‹Â

Â Â Â Â Â Â Â Â Â Â Â Â Â (ii)\( B_1 \)is linearly independent

Let â€‹\( T \in L(V, W) \)â€‹ be any vector. Then â€‹\( T(\alpha_1)\in W \)â€‹ and â€‹\( T(\alpha_1) \)â€‹ can be expressed as the linear combination of â€‹\( \beta_1, \beta_2, …, \beta_m \)â€‹.

i.e. â€‹

\[ T(\alpha_1)=a_{11}\beta_1+a_{21}\beta_2+ … +a_{m1}\beta_m, \ where \ a_{11}, a_{21}, … , a_{m1}\in F \]

â€‹i.e. â€‹\( T(\alpha_1)=\displaystyle\sum_{p=1}^{m}a_{p1}\beta_p \)â€‹Â

Now for each i, â€‹\( 1 \le i \le n \)â€‹,â€‹

\[ T(\alpha_i)=a_{1i}\beta_1+a_{2i}\beta_2+ … +a_{mi}\beta_mÂ Â T(\alpha_i)=\displaystyle\sum_{p=1}^{m}a_{pi}\beta_p \ — \ (1);\ 1 \le i \le n \]

â€‹Â Let â€‹\( S=\displaystyle\sum_{p=1}^{m}\displaystyle\sum_{q=1}^{n}a_{pq}T_{pq} \)â€‹Â

Then, â€‹\( S\in L(V, W) \)â€‹Â

Claim that â€‹\( S=T \)â€‹Â

Consider, â€‹\( S(\alpha_i) \)â€‹

â€‹\( =\left(\displaystyle\sum_{p=1}^{m}\displaystyle\sum_{q=1}^{n}a_{pq}T_{pq}\right)(\alpha_i) \)â€‹Â

â€‹\( =\displaystyle\sum_{p=1}^{m}\left(\displaystyle\sum_{q=1}^{n}a_{pq}T_{pq}(\alpha_i)\right) \)â€‹

â€‹\( =\displaystyle\sum_{p=1}^{m}\left(\displaystyle\sum_{q=1}^{n}a_{pq}\delta_{iq}\beta_p\right) \)â€‹

\( =\displaystyle\sum_{p=1}^{m}a_{pi}\beta_p \)Â

â€‹\( =T(\alpha_i) \)â€‹

Then â€‹\( S(\alpha_i)=T(\alpha_i) \)â€‹

This shows that â€‹\( S=T \)â€‹.

hence, â€‹\( L(B_1)=L(V, W) \)â€‹.

Now to show that â€‹\( B_1 \)â€‹ is linearly independent.

Suppose that â€‹\( \displaystyle\sum_{p=1}^{m}\displaystyle\sum_{q=1}^{n}b_{pq}T_{pq}=0 \)â€‹

â€‹\( \implies \left(\displaystyle\sum_{p=1}^{m}\displaystyle\sum_{q=1}^{n}b_{pq}T_{pq}\right)(\alpha_i)=0(\alpha_i)=0 \)â€‹

â€‹\( \implies \displaystyle\sum_{p=1}^{m}\left(\displaystyle\sum_{q=1}^{n}b_{pq}T_{pq}(\alpha_i)\right)=0 \)â€‹

â€‹\( \implies \displaystyle\sum_{p=1}^{m}\left(\displaystyle\sum_{q=1}^{n}b_{pq}\delta_{iq}\beta_p)\right)=0 \)â€‹

â€‹\( \implies \displaystyle\sum_{p=1}^{m}b_{pi}\beta_p=0 \)â€‹

â€‹\( \because \{\beta_1, \beta_2, … ,\beta_m\} \)â€‹Â is linearly independent,

â€‹\( \implies b_{pq}=0 \)â€‹ for â€‹\( 1 \le p \le m, 1 \le q \le n \)â€‹.Â

â€‹\( \therefore B_1 \)â€‹Â is linearly independent.Â

â€‹\( \therefore B_1 \)â€‹ forms basis for â€‹\( L(V, W) \ and \ \# B_1=mn \)â€‹.Â

Hence, â€‹\( B_1 \)â€‹ is finite dimensional and has dimension mn.

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