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HomeMathematicsAlgebraLinear Transformation Part I - Algebra of Linear Transformation

Linear Transformation Part I – Algebra of Linear Transformation

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Theorem: 

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

Proof:

Let ​\( \alpha, \beta \in V \)​ and c \in F and S & T are linear transformation from V to W. 
\( (S+T)(c\alpha+\beta) \)
\( =S(c\alpha+\beta)+T(c\alpha+\beta) \)
\( =cS(\alpha)+S(\beta)+cT(\alpha)+T(\beta) \)
\( =c(S+T)(\alpha)+(S+T)(\alpha) \)​ 
 which shows that ​\( (S+T) \)​ is a linear transformation. 

Similarly, ​

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

\( =cd(T(\alpha))+cT(\beta) \)
\( =d[cT(\alpha)]+cT(\alpha) \)
\( =d[(cT)(\alpha)]+(cT)(\beta) \)
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 ​\( L(V, W) \)​. Thus, ​\( L(V, W)=\{T:V\rightarrow W| T \ is \ a \ L.T.\} \)​ is a vector space over the field F w.r.t. 
\( (S+T)(\alpha)=S(\alpha)+T(\alpha) \)​.
\( (cT)(\alpha)=c(T(\alpha)) \)
\( \forall S, T \in L(V, W), \alpha \in V \)​ and ​\( c \in F \)​.
 

Theorem:

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

Proof: 

Let ​\( B=\{\alpha_1, \alpha_2, … ,\alpha_n\} \)​ and ​\( B’=\{\beta_1, \beta_2, … ,\beta_m\} \)​ be ordered basis for V and W respectively.
Then by theorem, there exists a unique linear transformation ​\( T_{11} \)​ such that

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

where ​\( \beta_1, 0, 0, … , 0 \)​ are the vectors in W.
But for any pair ​\( (p, q) \)​ where ​\( 1\le p \le m \)​ and ​\( 1 \le q \le n \)​ there exists a linear transformation ​\( T_{pq} \)​ from V into W such that ​\( T_{pq} = \begin{cases} 0 & {if \ \ i \ne q} \\ \beta_p & {if \ \ i = q} \end{cases} \)
i.e. ​\( T_{pq}(\alpha_i)=\delta_{iq}\beta_p, \text{where} \ \delta_{iq} = \begin{cases} 1 & {if \ \ i=q} \\ 0 & {if \ \ i \ne q} \end{cases} \)
Since, ​\( 1\le p \le m \)​ and ​\( 1 \le q \le n \)​, there are mn such\( T_{pq} \)‘s. 

Let ​

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

Claim that ​\( B_1 \)​ is the basis for ​\( L(V, W) \)
i.e. to show that (i) ​\( L(B_1)=L(V, W) \)​ 
                          (ii)\( B_1 \)is linearly independent
Let ​\( T \in L(V, W) \)​ be any vector. Then ​\( T(\alpha_1)\in W \)​ and ​\( T(\alpha_1) \)​ can be expressed as the linear combination of ​\( \beta_1, \beta_2, …, \beta_m \)​.

i.e. ​

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

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

Now for each i, ​\( 1 \le i \le n \)​,

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

​ Let ​\( S=\displaystyle\sum_{p=1}^{m}\displaystyle\sum_{q=1}^{n}a_{pq}T_{pq} \)​ 

Then, ​\( S\in L(V, W) \)​ 
Claim that ​\( S=T \)​ 
Consider, ​\( S(\alpha_i) \)
\( =\left(\displaystyle\sum_{p=1}^{m}\displaystyle\sum_{q=1}^{n}a_{pq}T_{pq}\right)(\alpha_i) \)​ 
\( =\displaystyle\sum_{p=1}^{m}\left(\displaystyle\sum_{q=1}^{n}a_{pq}T_{pq}(\alpha_i)\right) \)
\( =\displaystyle\sum_{p=1}^{m}\left(\displaystyle\sum_{q=1}^{n}a_{pq}\delta_{iq}\beta_p\right) \)
\( =\displaystyle\sum_{p=1}^{m}a_{pi}\beta_p \) 
\( =T(\alpha_i) \)
Then ​\( S(\alpha_i)=T(\alpha_i) \)
This shows that ​\( S=T \)​.
hence, ​\( L(B_1)=L(V, W) \)​.
Now to show that ​\( B_1 \)​ is linearly independent.
Suppose that ​\( \displaystyle\sum_{p=1}^{m}\displaystyle\sum_{q=1}^{n}b_{pq}T_{pq}=0 \)
\( \implies \left(\displaystyle\sum_{p=1}^{m}\displaystyle\sum_{q=1}^{n}b_{pq}T_{pq}\right)(\alpha_i)=0(\alpha_i)=0 \)
\( \implies \displaystyle\sum_{p=1}^{m}\left(\displaystyle\sum_{q=1}^{n}b_{pq}T_{pq}(\alpha_i)\right)=0 \)
\( \implies \displaystyle\sum_{p=1}^{m}\left(\displaystyle\sum_{q=1}^{n}b_{pq}\delta_{iq}\beta_p)\right)=0 \)
\( \implies \displaystyle\sum_{p=1}^{m}b_{pi}\beta_p=0 \)
\( \because \{\beta_1, \beta_2, … ,\beta_m\} \)​ is linearly independent,
\( \implies b_{pq}=0 \)​ for ​\( 1 \le p \le m, 1 \le q \le n \)​. 
\( \therefore B_1 \)​ is linearly independent. 
\( \therefore B_1 \)​ forms basis for ​\( L(V, W) \ and \ \# B_1=mn \)​. 
Hence, ​\( B_1 \)​ is finite dimensional and has dimension mn.
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