Editing Change of basis (section)

==In terms of linear maps==
Normally, a [[matrix (mathematics)|matrix]] represents a [[linear map]], and the product of a matrix and a column vector represents the [[function application]] of the corresponding linear map to the vector whose coordinates form the column vector. The change-of-basis formula is a specific case of this general principle, although this is not immediately clear from its definition and proof.

When one says that a matrix ''represents'' a linear map, one refers implicitly to [[basis (linear algebra)|bases]] of implied vector spaces, and to the fact that the choice of a basis induces an [[linear isomorphism|isomorphism]] between a vector space and {{math|''F''{{sup|''n''}}}}, where {{mvar|F}} is the field of scalars. When only one basis is considered for each vector space, it is worth to leave this isomorphism implicit, and to work [[up to]] an isomorphism. As several bases of the same vector space are considered here, a more accurate wording is required.

Let {{mvar|F}} be a [[field (mathematics)|field]], the set <math>F^n</math> of the [[tuple|{{mvar|n}}-tuples]] is a {{mvar|F}}-vector space whose addition and scalar multiplication are defined component-wise. Its [[standard basis]] is the basis that has as its {{mvar|i}}th element the tuple with all components equal to {{math|0}} except the {{mvar|i}}th that is {{math|1}}.

A basis <math>B=(v_1, \ldots, v_n)</math> of a {{mvar|F}}-vector space {{mvar|V}} defines a [[linear isomorphism]] <math>\phi\colon F^n\to V</math> by 
:<math>\phi(x_1,\ldots,x_n)=\sum_{i=1}^n x_i v_i.</math>
Conversely, such a linear isomorphism defines a basis, which is the image by <math>\phi</math> of the standard basis of <math>F^n.</math>

Let <math>B_\mathrm {old}=(v_1, \ldots, v_n)</math> be the "old basis" of a change of basis, and <math>\phi_\mathrm {old}</math> the associated isomorphism. Given a change-of basis matrix {{mvar|A}}, one could consider it the matrix of an [[endomorphism]] <math>\psi_A</math> of <math>F^n.</math> Finally, define
:<math>\phi_\mathrm{new}=\phi_\mathrm{old}\circ\psi_A</math> 
(where <math>\circ</math> denotes [[function composition]]), and
:<math>B_\mathrm{new}= \phi_\mathrm{new}(\phi_\mathrm{old}^{-1}(B_\mathrm{old})). </math>
A straightforward verification shows that this definition of <math>B_\mathrm{new}</math> is the same as that of the preceding section.

Now, by composing the equation <math>\phi_\mathrm{new}=\phi_\mathrm{old}\circ\psi_A</math> with <math>\phi_\mathrm{old}^{-1}</math> on the left and <math>\phi_\mathrm{new}^{-1}</math> on the right, one gets 
:<math>\phi_\mathrm{old}^{-1} = \psi_A \circ \phi_\mathrm{new}^{-1}.</math>
It follows that, for <math>v\in V,</math> one has 
:<math>\phi_\mathrm{old}^{-1}(v)= \psi_A(\phi_\mathrm{new}^{-1}(v)),</math>
which is the change-of-basis formula expressed in terms of linear maps instead of coordinates.