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Matrix Representation of Linear Operators

Posted November 16th, 2007 by Structure

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

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

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

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

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}<br /> a_{1,1}&a_{1,2}&.&a_{1,n}\\</p> <p>a_{2,1}&a_{2,1}&.&a_{2,n}\\<br /> .&.&.&..\\<br /> a_{m,1}&a_{m,2}&.&a_{m,n}<br /> \end {array}\right ] \)<br /> \left [\begin{array}{c}<br /> x_1\\<br /> x_2\\...\\x_n\end {array}\right ] \)=\left [\begin{array}{c}<br /> y_1\\y_2\\.\\y_m\end {array}\right ] \) $

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Comments

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Very useful article.

On February 23rd, 2011 alpha says:

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

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