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HomeMathematicsAlgebraLinear Transformation Part I - Algebra of Linear Transformation

# Linear Transformation Part I – 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 ​$$T_{pq} = \begin{cases} 0 & {if \ \ i \ne q} \\ \beta_p & {if \ \ i = q} \end{cases}$$
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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