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HomeMathematicsAlgebraLinear Transformation Part I - Algebra of Linear Transformation

# Linear Transformation Part I – Algebra of Linear Transformation

### Theorem:Â

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

### Proof:

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

Similarly, â€‹

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

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

### Theorem:

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

### Proof:Â

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

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

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

Let â€‹

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

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

i.e. â€‹

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

â€‹i.e. â€‹$$T(\alpha_1)=\displaystyle\sum_{p=1}^{m}a_{p1}\beta_p$$â€‹Â

Now for each i, â€‹$$1 \le i \le n$$â€‹,â€‹

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

â€‹Â Let â€‹$$S=\displaystyle\sum_{p=1}^{m}\displaystyle\sum_{q=1}^{n}a_{pq}T_{pq}$$â€‹Â

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