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HomeMathematicsAlgebraLinear Transformation Part I - Algebra of Linear Transformation

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 $W$. 
$(S+T)(calpha+beta)=S(calpha+beta)+T(calpha+beta)$ 
                              $=cS(alpha)+S(beta)+cT(alpha)+T(beta)$  
                              $=c(S+T)(alpha)+(S+T)(alpha)$ 
 which shows that $(S+T)$ is a linear transformation. 
Similarly, $(cT)(dalpha+beta)=c[T(dalpha+beta)]=c[dT(alpha)+T(beta)]$ 
$=cd(T(alpha))+cT(beta)$ 
$=d[cT(alpha)]+cT(alpha)$ 
$=d[(cT)(alpha)]+(cT)(beta)$ 
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, $L(V, W)={T:Vrightarrow W| T is a L.T.}$ 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 $1le p le m$ 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, $1le p le m$ and $1 le q le n$, there are $mn$ such $T_{pq}$’s. 
Let $B_1={T_{pq}|1le p le m and 1 le q le n}$ $(# B_1=mn)$ 
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. $T(alpha_1)=displaystylesum_{p=1}^{m}a_{p1}beta_p$ 
Now for each i, $1 le i le n$,
$T(alpha_i)=a_{1i}beta_1+a_{2i}beta_2+ … +a_{mi}beta_m$ 
 $T(alpha_i)=displaystylesum_{p=1}^{m}a_{pi}beta_p$ ——– (1) $1 le i le n$ 
Let $S=displaystylesum_{p=1}^{m}displaystylesum_{q=1}^{n}a_{pq}T_{pq}$ 
Then, $Sin L(V, W)$ 
Claim that $S=T$ 

Consider, $S(alpha_i)$
$=left(displaystylesum_{p=1}^{m}displaystylesum_{q=1}^{n}a_{pq}T_{pq}right)(alpha_i)$ 
$=displaystylesum_{p=1}^{m}left(displaystylesum_{q=1}^{n}a_{pq}T_{pq}(alpha_i)right)$ 
$=displaystylesum_{p=1}^{m}left(displaystylesum_{q=1}^{n}a_{pq}delta_{iq}beta_pright)$ 
$=displaystylesum_{p=1}^{m}a_{pi}beta_p$ 
$=T(alpha_i)$ 
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 $displaystylesum_{p=1}^{m}displaystylesum_{q=1}^{n}b_{pq}T_{pq}=0$ 
$implies left(displaystylesum_{p=1}^{m}displaystylesum_{q=1}^{n}b_{pq}T_{pq}right)(alpha_i)=0(alpha_i)=0$ 
$implies displaystylesum_{p=1}^{m}left(displaystylesum_{q=1}^{n}b_{pq}T_{pq}(alpha_i)right)=0$ 
$implies displaystylesum_{p=1}^{m}left(displaystylesum_{q=1}^{n}b_{pq}delta_{iq}beta_p)right)=0$ 
$implies displaystylesum_{p=1}^{m}b_{pi}beta_p=0$ 
$because {beta_1, beta_2, … ,beta_m}$ 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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