Change of basis

Not to be confused withChange of base.

Inmathematics,anordered basisof avector spaceof finitedimensionnallows representing uniquely any element of the vector space by acoordinate vector,which is asequenceofnscalarscalledcoordinates.If two different bases are considered, the coordinate vector that represents a vectorvon one basis is, in general, different from the coordinate vector that representsvon the other basis. Achange of basisconsists of converting every assertion expressed in terms of coordinates relative to one basis into an assertion expressed in terms of coordinates relative to the other basis.

Alinear combinationof one basis of vectors (purple) obtains new vectors (red). If they arelinearly independent,these form a new basis. The linear combinations relating the first basis to the other extend to alinear transformation,called the change of basis.

A vector represented by two different bases (purple and red arrows).

Such a conversion results from thechange-of-basis formulawhich expresses the coordinates relative to one basis in terms of coordinates relative to the other basis. Usingmatrices,this formula can be written

$\mathbf {x} _{\mathrm {old} }=A\,\mathbf {x} _{\mathrm {new} },$

where "old" and "new" refer respectively to the initially defined basis and the other basis,$\mathbf {x} _{\mathrm {old} }$and$\mathbf {x} _{\mathrm {new} }$are thecolumn vectorsof the coordinates of the same vector on the two bases.$A$is thechange-of-basis matrix(also calledtransition matrix), which is the matrix whose columns are the coordinates of the newbasis vectorson the old basis.

A change of basis is sometimes called achange of coordinates,although it excludes manycoordinate transformations. For applications inphysicsand specially inmechanics,a change of basis often involves the transformation of anorthonormal basis,understood as arotationinphysical space,thus excludingtranslations. This article deals mainly with finite-dimensional vector spaces. However, many of the principles are also valid for infinite-dimensional vector spaces.

Change of basis formula

Let$B_{\mathrm {old} }=(v_{1},\ldots,v_{n})$be a basis of afinite-dimensional vector spaceVover afieldF.

Forj= 1,...,n,one can define a vectorw_jby its coordinates$a_{i,j}$over$B_{\mathrm {old} }\colon$

$w_{j}=\sum _{i=1}^{n}a_{i,j}v_{i}.$

Let

$A=\left(a_{i,j}\right)_{i,j}$

be thematrixwhosejth column is formed by the coordinates ofw_j.(Here and in what follows, the indexirefers always to the rows ofAand the$v_{i},$while the indexjrefers always to the columns ofAand the$w_{j};$such a convention is useful for avoiding errors in explicit computations.)

Setting$B_{\mathrm {new} }=(w_{1},\ldots,w_{n}),$one has that$B_{\mathrm {new} }$is a basis ofVif and only if the matrixAisinvertible,or equivalently if it has a nonzerodeterminant.In this case,Ais said to be thechange-of-basis matrixfrom the basis$B_{\mathrm {old} }$to the basis$B_{\mathrm {new} }.$

Given a vector$z\in V,$let$(x_{1},\ldots,x_{n})$be the coordinates of$z$over$B_{\mathrm {old} },$and$(y_{1},\ldots,y_{n})$its coordinates over$B_{\mathrm {new} };$that is

$z=\sum _{i=1}^{n}x_{i}v_{i}=\sum _{j=1}^{n}y_{j}w_{j}.$

(One could take the same summation index for the two sums, but choosing systematically the indexesifor the old basis andjfor the new one makes clearer the formulas that follows, and helps avoiding errors in proofs and explicit computations.)

Thechange-of-basis formulaexpresses the coordinates over the old basis in terms of the coordinates over the new basis. With above notation, it is

$x_{i}=\sum _{j=1}^{n}a_{i,j}y_{j}\qquad {\text{for }}i=1,\ldots,n.$

In terms of matrices, the change of basis formula is

$\mathbf {x} =A\,\mathbf {y},$

where$\mathbf {x}$and$\mathbf {y}$are the column vectors of the coordinates ofzover$B_{\mathrm {old} }$and$B_{\mathrm {new} },$respectively.

Proof:Using the above definition of the change-of basis matrix, one has

${\begin{aligned}z&=\sum _{j=1}^{n}y_{j}w_{j}\\&=\sum _{j=1}^{n}\left(y_{j}\sum _{i=1}^{n}a_{i,j}v_{i}\right)\\&=\sum _{i=1}^{n}\left(\sum _{j=1}^{n}a_{i,j}y_{j}\right)v_{i}.\end{aligned}}$

As$z=\textstyle \sum _{i=1}^{n}x_{i}v_{i},$the change-of-basis formula results from the uniqueness of the decomposition of a vector over a basis.

Example

Consider theEuclidean vector space$\mathbb {R} ^{2}$and a basis consisting of the vectors$v_{1}=(1,0)$and$v_{2}=(0,1).$If onerotatesthem by an angle oft,one has anew basisformed by$w_{1}=(\cos t,\sin t)$and$w_{2}=(-\sin t,\cos t).$

So, the change-of-basis matrix is ${\begin{bmatrix}\cos t&-\sin t\\\sin t&\cos t\end{bmatrix}}.$

The change-of-basis formula asserts that, if$y_{1},y_{2}$are the new coordinates of a vector$(x_{1},x_{2}),$then one has

${\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}={\begin{bmatrix}\cos t&-\sin t\\\sin t&\cos t\end{bmatrix}}\,{\begin{bmatrix}y_{1}\\y_{2}\end{bmatrix}}.$

That is,

$x_{1}=y_{1}\cos t-y_{2}\sin t\qquad {\text{and}}\qquad x_{2}=y_{1}\sin t+y_{2}\cos t.$

This may be verified by writing

${\begin{aligned}x_{1}v_{1}+x_{2}v_{2}&=(y_{1}\cos t-y_{2}\sin t)v_{1}+(y_{1}\sin t+y_{2}\cos t)v_{2}\\&=y_{1}(\cos(t)v_{1}+\sin(t)v_{2})+y_{2}(-\sin(t)v_{1}+\cos(t)v_{2})\\&=y_{1}w_{1}+y_{2}w_{2}.\end{aligned}}$

In terms of linear maps

Normally, amatrixrepresents alinear map,and the product of a matrix and a column vector represents thefunction applicationof 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 matrixrepresentsa linear map, one refers implicitly tobasesof implied vector spaces, and to the fact that the choice of a basis induces anisomorphismbetween a vector space andFⁿ,whereFis the field of scalars. When only one basis is considered for each vector space, it is worth to leave this isomorphism implicit, and to workup toan isomorphism. As several bases of the same vector space are considered here, a more accurate wording is required.

LetFbe afield,the set$F^{n}$of then-tuplesis aF-vector space whose addition and scalar multiplication are defined component-wise. Itsstandard basisis the basis that has as itsith element the tuple with all components equal to0except theith that is1.

A basis$B=(v_{1},\ldots,v_{n})$of aF-vector spaceVdefines alinear isomorphism$\phi \colon F^{n}\to V$by

$\phi (x_{1},\ldots,x_{n})=\sum _{i=1}^{n}x_{i}v_{i}.$

Conversely, such a linear isomorphism defines a basis, which is the image by$\phi$of the standard basis of$F^{n}.$

Let$B_{\mathrm {old} }=(v_{1},\ldots,v_{n})$be the "old basis" of a change of basis, and$\phi _{\mathrm {old} }$the associated isomorphism. Given a change-of basis matrixA,one could consider it the matrix of anendomorphism$\psi _{A}$of$F^{n}.$Finally, define

$\phi _{\mathrm {new} }=\phi _{\mathrm {old} }\circ \psi _{A}$

(where$\circ$denotesfunction composition), and

$B_{\mathrm {new} }=\phi _{\mathrm {new} }(\phi _{\mathrm {old} }^{-1}(B_{\mathrm {old} })).$

A straightforward verification shows that this definition of$B_{\mathrm {new} }$is the same as that of the preceding section.

Now, by composing the equation$\phi _{\mathrm {new} }=\phi _{\mathrm {old} }\circ \psi _{A}$with$\phi _{\mathrm {old} }^{-1}$on the left and$\phi _{\mathrm {new} }^{-1}$on the right, one gets

$\phi _{\mathrm {old} }^{-1}=\psi _{A}\circ \phi _{\mathrm {new} }^{-1}.$

It follows that, for$v\in V,$one has

$\phi _{\mathrm {old} }^{-1}(v)=\psi _{A}(\phi _{\mathrm {new} }^{-1}(v)),$

which is the change-of-basis formula expressed in terms of linear maps instead of coordinates.

Function defined on a vector space

Afunctionthat has a vector space as itsdomainis commonly specified as amultivariate functionwhose variables are the coordinates on some basis of the vector on which the function isapplied.

When the basis is changed, theexpressionof the function is changed. This change can be computed by substituting the "old" coordinates for their expressions in terms of the "new" coordinates. More precisely, iff(x)is the expression of the function in terms of the old coordinates, and ifx=Ayis the change-of-base formula, thenf(Ay)is the expression of the same function in terms of the new coordinates.

The fact that the change-of-basis formula expresses the old coordinates in terms of the new one may seem unnatural, but appears as useful, as nomatrix inversionis needed here.

As the change-of-basis formula involves onlylinear functions,many function properties are kept by a change of basis. This allows defining these properties as properties of functions of a variable vector that are not related to any specific basis. So, a function whose domain is a vector space or a subset of it is

• a linear function,
• apolynomial function,
• acontinuous function,
• adifferentiable function,
• asmooth function,
• ananalytic function,

if the multivariate function that represents it on some basis—and thus on every basis—has the same property.

This is specially useful in the theory ofmanifolds,as this allows extending the concepts of continuous, differentiable, smooth and analytic functions to functions that are defined on a manifold.

Linear maps

Consider alinear mapT:W→Vfrom avector spaceWof dimensionnto a vector spaceVof dimensionm.It is represented on "old" bases ofVandWby am×nmatrixM.A change of bases is defined by anm×mchange-of-basis matrixPforV,and ann×nchange-of-basis matrixQforW.

On the "new" bases, the matrix ofTis

$P^{-1}MQ.$

This is a straightforward consequence of the change-of-basis formula.

Endomorphisms

Endomorphismsare linear maps from a vector spaceVto itself. For a change of basis, the formula of the preceding section applies, with the same change-of-basis matrix on both sides of the formula. That is, ifMis thesquare matrixof an endomorphism ofVover an "old" basis, andPis a change-of-basis matrix, then the matrix of the endomorphism on the "new" basis is

$P^{-1}MP.$

As everyinvertible matrixcan be used as a change-of-basis matrix, this implies that two matrices aresimilarif and only if they represent the same endomorphism on two different bases.

Bilinear forms

Abilinear formon a vector spaceVover afieldFis a functionV×V→ Fwhich islinearin both arguments. That is,B:V×V→ Fis bilinear if the maps $v\mapsto B(v,w)$ and$v\mapsto B(w,v)$ are linear for every fixed$w\in V.$

The matrixBof a bilinear formBon a basis$(v_{1},\ldots,v_{n})$(the "old" basis in what follows) is the matrix whose entry of theith row andjth column is$B(v_{i},v_{j})$.It follows that ifvandware the column vectors of the coordinates of two vectorsvandw,one has

$B(v,w)=\mathbf {v} ^{\mathsf {T}}\mathbf {B} \mathbf {w},$

where$\mathbf {v} ^{\mathsf {T}}$denotes thetransposeof the matrixv.

IfPis a change of basis matrix, then a straightforward computation shows that the matrix of the bilinear form on the new basis is

$P^{\mathsf {T}}\mathbf {B} P.$

Asymmetric bilinear formis a bilinear formBsuch that$B(v,w)=B(w,v)$for everyvandwinV.It follows that the matrix ofBon any basis issymmetric.This implies that the property of being a symmetric matrix must be kept by the above change-of-base formula. One can also check this by noting that the transpose of a matrix product is the product of the transposes computed in the reverse order. In particular,

$(P^{\mathsf {T}}\mathbf {B} P)^{\mathsf {T}}=P^{\mathsf {T}}\mathbf {B} ^{\mathsf {T}}P,$

and the two members of this equation equal$P^{\mathsf {T}}\mathbf {B} P$if the matrixBis symmetric.

If thecharacteristicof the ground fieldFis not two, then for every symmetric bilinear form there is a basis for which the matrix isdiagonal.Moreover, the resulting nonzero entries on the diagonal are defined up to the multiplication by a square. So, if the ground field is the field$\mathbb {R}$of thereal numbers,these nonzero entries can be chosen to be either1or–1.Sylvester's law of inertiais a theorem that asserts that the numbers of1and of–1depends only on the bilinear form, and not of the change of basis.

Symmetric bilinear forms over the reals are often encountered ingeometryandphysics,typically in the study ofquadricsand of theinertiaof arigid body.In these cases,orthonormal basesare specially useful; this means that one generally prefer to restrict changes of basis to those that have anorthogonalchange-of-base matrix, that is, a matrix such that$P^{\mathsf {T}}=P^{-1}.$Such matrices have the fundamental property that the change-of-base formula is the same for a symmetric bilinear form and the endomorphism that is represented by the same symmetric matrix. TheSpectral theoremasserts that, given such a symmetric matrix, there is an orthogonal change of basis such that the resulting matrix (of both the bilinear form and the endomorphism) is a diagonal matrix with theeigenvaluesof the initial matrix on the diagonal. It follows that, over the reals, if the matrix of an endomorphism is symmetric, then it isdiagonalizable.

See also

• Active and passive transformation
• Covariance and contravariance of vectors
• Integral transform,the continuous analogue of change of basis.
• Chirgwin-Coulson weights— application in computational chemistry

Bibliography

• Anton, Howard (1987),Elementary Linear Algebra(5th ed.), New York:Wiley,ISBN0-471-84819-0
• Beauregard, Raymond A.; Fraleigh, John B. (1973),A First Course In Linear Algebra: with Optional Introduction to Groups, Rings, and Fields,Boston:Houghton Mifflin Company,ISBN0-395-14017-X
• Nering, Evar D. (1970),Linear Algebra and Matrix Theory(2nd ed.), New York:Wiley,LCCN76091646

External links

• MIT Linear Algebra Lecture on Change of Basis,from MIT OpenCourseWare
• Khan Academy Lecture on Change of Basis,from Khan Academy