Theorem (1):
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 \( c \in F \).
Also Read: Linear Transformation Part II – Inverse Linear Transformation and Isomorphism
Theorem (2):
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.
