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Changing basis

Posted November 16th, 2007 by Structure

We are interested in the form of the matrix associated to a linear operator after a basis' change.
Let $ (V,+,.,k) $ be a n dimensional k-vector space over a field k. Let $ \mathcal{B}=\{e_1,e_2,...e_n\} $ and $ \mathcal{B'}=\{e'_1,e'_2,...e'_n\} $ be two basis of V. We shall call transition matrix from $ \mathcal{B} $ to $ \mathcal{B'} $ the matrix associated to identity map of V in the two basis $ \mathcal{B'} $ and $ \mathcal{B} $ namely $ T=\mathcal{M}(id_{V},\mathcal{B'},\mathcal{B}) $
After the previous definition we have $ e'_j=\sum_{i=1}^{n}t_{i,j}e_i $. Let $ x=\sum_{i=1}^{n}x_i e_{i}=\sum_{j=1}^{n}x'_j e'_{j} $.
Then $ x=\sum_{j=1}^{n}x'_j e'_{j}=\sum_{j=1}^{n}x'_j \sum_{i=1}^{n}t_{i,j}e_i=\sum_{i=1}^{n}(\sum_{j=1}^{n}t_{i,j}x'_j) x_i=\sum_{i=1}^{n}x_i e_{i} $
so we have for all $ 1\leq i \leq n\; $
$ x_i=\sum_{j=1}^{n}t_{i,j}x'_j $ or in matrix form

$$X=TX'.$$

$ \left [\begin{array}{c}<br /> x_1\\<br /> x_2\\.\\x_n\end {array}\right ] \)=\left [\begin{array}{cccc}<br /> t_{1,1}&t_{1,2}&.&t_{1,n}\\</p> <p>t_{2,1}&t_{2,1}&.&t_{2,n}\\<br /> .&.&.&..\\<br /> t_{n,1}&t_{n,2}&.&t_{n,n}<br /> \end {array}\right ] \)<br /> \left [\begin{array}{c}<br /> x'_1\\<br /> x'_2\\...\\x'_n\end {array}\right ] \) $
Now it is easy to find what relation is between associated matrix.
Let $ \mathcal{A}\in\mathcal{L}_k(V_1,V_2) $ ,$ \mathcal{B}_1,\mathcal{B'}_1 $ two basis in $ V_1 $, $ T=\mathcal{M}(id_{{V}_1},\mathcal{B'}_1,\mathcal{B}_1) $
$ \mathcal{B}_2,\mathcal{B'}_2 $ two basis in $ V_2 $, $ A=\mathcal{M}(\mathcal{A},\mathcal{B}_1,\mathcal{B}_2) $
$ S=\mathcal{M}(id_{{V}_2},\mathcal{B'}_2,\mathcal{B}_2) $. Then $ S^{-1}=\mathcal{M}(id_{{V}_2},\mathcal{B}_2,\mathcal{B'}_2) $.
Now $ A'=\mathcal{M}(\mathcal{A},\mathcal{B'}_1,\mathcal{B'}_2)=\mathcal{M}(id_{{V}_2},\mathcal{B}_2,\mathcal{B'}_2<br /> )\mathcal{M}(\mathcal{A},\mathcal{B}_1,\mathcal{B}_2)\mathcal{M}(id_{{V}_1},\mathcal{B'}_1,\mathcal{B}_1) $

Finally

$$A'=S^{-1}AT$$

For those familiar with manifold, now or in the future, this is a very elementary example of local charts and maps and how they described a manifold morphism.
A vector space can be viewed as an abstract "manifold ". For a basis, the correspondence $ h_{V}(x)=X\in R^n $ is a global map , linear transform $ \mathcal{A}:V_1\rightarrow V_2 $ is an application and $ A= \mathcal{M}(\mathcal{A},\mathcal{B}_1,\mathcal{B}_2) $ is its expression in the local charts.
$ A(X)=h_{V_2}\mathcal{A}h^{-1}_{V_1}(X)=h_{V_2}\mathcal{A}(x)=h_{V_2}(y)=Y $
Relation

$$A'=S^{-1}AT$$

shows how the linear map looks like in another local system.

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