Linear Transformation Part I - Algebra of Linear Transformation

Algebra of Linear Transformation

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