Matrix Representation of Linear Operators

Submitted by Structure on Fri, 11/16/2007 - 08:54.

Let us consider a k-vector space morphism of vector space between two finite dimensional k-vector spaces.over a field k. We shall call such an object linear transform or linear operator.
So let $V_1,V_2$ be two k-vector and $\mathcal{L}_k(V_1,V_2)$ the set of k-linear operators from $V_1$ to $V_2$ . If $\mathcal{A}\in\mathcal{L}_k(V_1,V_2)$ then $\forall\lambda,\mu\in k,x,y\in V_1$ we have

$\mathcal{A}(\lambda x+\mu y)=\lambda \mathcal{A}(x)+\mu \mathcal{A}(y)$

Let $\mathcal{B}_1=\{e_1,e_2,...e_n\}$ be a basis in $V_1$ and $\mathcal{B}_2=\{u_1,u_2,...u_m\}$ be a basis in $V_2$ . For all $x\in V_1$ there are $x_i\in k,i=1,...n$ such as $x=\sum_{j=1}^{n}x_je_j$ .
Now

$\mathcal{A}(x)=\mathcal{A}(\sum_{j=1}^{n}x_je_j)=\sum_{j=1}^{n}x_j\mathcal{A}(e_j)=\sum_{j=1}^{n}x_j\sum_{i=1}^{m}a_{i,j}u_i =\sum_{i=1}^{m}(\sum_{j=1}^{n}a_{i,j}x_j)u_i$

But if we write $\mathcal{A}(x)=y=\sum_{i=1}^{m}y_iu_i$ we have

$\mathcal{A}(x)=\mathcal{A}(\sum_{j=1}^{n}x_je_j)=\sum_{j=1}^{n}x_j\mathcal{A}(e_j)=\sum_{j=1}^{n}x_j\sum_{i=1}^{m}a_{i,j}u_i =\sum_{i=1}^{m}(\sum_{j=1}^{n}a_{i,j}x_j)u_i=\sum_{i=1}^{m}y_iu_i$

and , as $u_i$ form a basis in $V_2$ we have for all i from 1 to m

$(1)\:\sum_{j=1}^{n}a_{i,j}x_j=y_i$

Let be $A=(a_{i,j})_{1\leq i\leq n,1\leq j \leq m}$ We shall call $A$ the matrix representation of $\mathcal{A}$ in the two basis $\mathcal{B}_1,\mathcal{B}_2$ and write $A=\mathcal{M}(\mathcal{A};\mathcal{B}_1,\mathcal{B}_2)$
Let $X=[x_1,x_2,....x_n]^A, Y=[y_1,y_2,....y_m]^A$ vectors in $R^n.R^m$ written in column form.
then relation

$\mathcal{A}(x)=y$

has an analog

$AX=Y$

deduced by (1)
$\left [\begin{array}{cccc} a_{1,1}&a_{1,2}&.&a_{1,n}\\ a_{2,1}&a_{2,1}&.&a_{2,n}\\ .&.&.&..\\ a_{m,1}&a_{m,2}&.&a_{m,n} \end {array}\right ] \) \left [\begin{array}{c} x_1\\ x_2\\...\\x_n\end {array}\right ] \)=\left [\begin{array}{c} y_1\\y_2\\.\\y_m\end {array}\right ] \)$

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Matrix Representation of Linear Operators