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Change of basis

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(Redirected fromChange of coordinates)
Coordinate change in linear algebra
Not to be confused withChange of base.
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Alinear combination of 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).

Inmathematics, anordered basis of avector space of finitedimensionn allows representing uniquely any element of the vector space by acoordinate vector, which is asequence ofnscalars calledcoordinates. If two different bases are considered, the coordinate vector that represents a vectorv on one basis is, in general, different from the coordinate vector that representsv on the other basis. Achange of basis consists 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.[1][2][3]

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

xold=Axnew,{\displaystyle \mathbf {x} _{\mathrm {old} }=A\,\mathbf {x} _{\mathrm {new} },}

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

A change of basis is sometimes called achange of coordinates, although it excludes manycoordinate transformations.For applications inphysics and specially inmechanics, a change of basis often involves the transformation of anorthonormal basis, understood as arotation inphysical 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

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LetBold=(v1,,vn){\displaystyle B_{\mathrm {old} }=(v_{1},\ldots ,v_{n})} be a basis of afinite-dimensional vector spaceV over afieldF.[a]

Forj = 1, ...,n, one can define a vectorwj by its coordinatesai,j{\displaystyle a_{i,j}} overBold:{\displaystyle B_{\mathrm {old} }\colon }

wj=i=1nai,jvi.{\displaystyle w_{j}=\sum _{i=1}^{n}a_{i,j}v_{i}.}

Let

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

be thematrix whosejth column is formed by the coordinates ofwj. (Here and in what follows, the indexi refers always to the rows ofA and thevi,{\displaystyle v_{i},} while the indexj refers always to the columns ofA and thewj;{\displaystyle w_{j};} such a convention is useful for avoiding errors in explicit computations.)

SettingBnew=(w1,,wn),{\displaystyle B_{\mathrm {new} }=(w_{1},\ldots ,w_{n}),} one has thatBnew{\displaystyle B_{\mathrm {new} }} is a basis ofV if and only if the matrixA isinvertible, or equivalently if it has a nonzerodeterminant. In this case,A is said to be thechange-of-basis matrix from the basisBold{\displaystyle B_{\mathrm {old} }} to the basisBnew.{\displaystyle B_{\mathrm {new} }.}

Given a vectorzV,{\displaystyle z\in V,} let(x1,,xn){\displaystyle (x_{1},\ldots ,x_{n})} be the coordinates ofz{\displaystyle z} overBold,{\displaystyle B_{\mathrm {old} },} and(y1,,yn){\displaystyle (y_{1},\ldots ,y_{n})} its coordinates overBnew;{\displaystyle B_{\mathrm {new} };} that is

z=i=1nxivi=j=1nyjwj.{\displaystyle 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 indexesi for the old basis andj for the new one makes clearer the formulas that follows, and helps avoiding errors in proofs and explicit computations.)

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

xi=j=1nai,jyjfor i=1,,n.{\displaystyle 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

x=Ay,{\displaystyle \mathbf {x} =A\,\mathbf {y} ,}

wherex{\displaystyle \mathbf {x} } andy{\displaystyle \mathbf {y} } are the column vectors of the coordinates ofz overBold{\displaystyle B_{\mathrm {old} }} andBnew,{\displaystyle B_{\mathrm {new} },} respectively.

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

z=j=1nyjwj=j=1n(yji=1nai,jvi)=i=1n(j=1nai,jyj)vi.{\displaystyle {\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}}}

Asz=i=1nxivi,{\displaystyle 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

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Consider theEuclidean vector spaceR2{\displaystyle \mathbb {R} ^{2}} and a basis consisting of the vectorsv1=(1,0){\displaystyle v_{1}=(1,0)} andv2=(0,1).{\displaystyle v_{2}=(0,1).} If onerotates them by an angle oft, one has anew basis formed byw1=(cost,sint){\displaystyle w_{1}=(\cos t,\sin t)} andw2=(sint,cost).{\displaystyle w_{2}=(-\sin t,\cos t).}

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

The change-of-basis formula asserts that, ify1,y2{\displaystyle y_{1},y_{2}} are the new coordinates of a vector(x1,x2),{\displaystyle (x_{1},x_{2}),} then one has

[x1x2]=[costsintsintcost][y1y2].{\displaystyle {\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,

x1=y1costy2sintandx2=y1sint+y2cost.{\displaystyle 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

x1v1+x2v2=(y1costy2sint)v1+(y1sint+y2cost)v2=y1(cos(t)v1+sin(t)v2)+y2(sin(t)v1+cos(t)v2)=y1w1+y2w2.{\displaystyle {\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

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Normally, amatrix represents alinear map, and the product of a matrix and a column vector represents thefunction 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 matrixrepresents a linear map, one refers implicitly tobases of implied vector spaces, and to the fact that the choice of a basis induces anisomorphism between a vector space andFn, whereF 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 workup to an isomorphism. As several bases of the same vector space are considered here, a more accurate wording is required.

LetF be afield, the setFn{\displaystyle F^{n}} of then-tuples is aF-vector space whose addition and scalar multiplication are defined component-wise. Itsstandard basis is the basis that has as itsith element the tuple with all components equal to0 except theith that is1.

A basisB=(v1,,vn){\displaystyle B=(v_{1},\ldots ,v_{n})} of aF-vector spaceV defines alinear isomorphismϕ:FnV{\displaystyle \phi \colon F^{n}\to V} by

ϕ(x1,,xn)=i=1nxivi.{\displaystyle \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 byϕ{\displaystyle \phi } of the standard basis ofFn.{\displaystyle F^{n}.}

LetBold=(v1,,vn){\displaystyle B_{\mathrm {old} }=(v_{1},\ldots ,v_{n})} be the "old basis" of a change of basis, andϕold{\displaystyle \phi _{\mathrm {old} }} the associated isomorphism. Given a change-of basis matrixA, one could consider it the matrix of anendomorphismψA{\displaystyle \psi _{A}} ofFn.{\displaystyle F^{n}.} Finally, define

ϕnew=ϕoldψA{\displaystyle \phi _{\mathrm {new} }=\phi _{\mathrm {old} }\circ \psi _{A}}

(where{\displaystyle \circ } denotesfunction composition), and

Bnew=ϕnew(ϕold1(Bold)).{\displaystyle B_{\mathrm {new} }=\phi _{\mathrm {new} }(\phi _{\mathrm {old} }^{-1}(B_{\mathrm {old} })).}

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

Now, by composing the equationϕnew=ϕoldψA{\displaystyle \phi _{\mathrm {new} }=\phi _{\mathrm {old} }\circ \psi _{A}} withϕold1{\displaystyle \phi _{\mathrm {old} }^{-1}} on the left andϕnew1{\displaystyle \phi _{\mathrm {new} }^{-1}} on the right, one gets

ϕold1=ψAϕnew1.{\displaystyle \phi _{\mathrm {old} }^{-1}=\psi _{A}\circ \phi _{\mathrm {new} }^{-1}.}

It follows that, forvV,{\displaystyle v\in V,} one has

ϕold1(v)=ψA(ϕnew1(v)),{\displaystyle \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

[edit]

Afunction that has a vector space as itsdomain is commonly specified as amultivariate function whose variables are the coordinates on some basis of the vector on which the function isapplied.

When the basis is changed, theexpression of 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 =Ay is 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 inversion is 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

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

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Consider alinear mapT:WV from avector spaceW of dimensionn to a vector spaceV of dimensionm. It is represented on "old" bases ofV andW by am×n matrixM. A change of bases is defined by anm×m change-of-basis matrixP forV, and ann×n change-of-basis matrixQ forW.

On the "new" bases, the matrix ofT is

P1MQ.{\displaystyle P^{-1}MQ.}

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

Endomorphisms

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Endomorphisms are linear maps from a vector spaceV to 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, ifM is thesquare matrix of an endomorphism ofV over an "old" basis, andP is a change-of-basis matrix, then the matrix of the endomorphism on the "new" basis is

P1MP.{\displaystyle P^{-1}MP.}

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

Bilinear forms

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Abilinear form on a vector spaceV over afieldF is a functionV ×V → F which islinear in both arguments. That is,B :V ×V → F is bilinear if the mapsvB(v,w){\displaystyle v\mapsto B(v,w)}andvB(w,v){\displaystyle v\mapsto B(w,v)}are linear for every fixedwV.{\displaystyle w\in V.}

The matrixB of a bilinear formB on a basis(v1,,vn){\displaystyle (v_{1},\ldots ,v_{n})} (the "old" basis in what follows) is the matrix whose entry of theith row andjth column isB(vi,vj){\displaystyle B(v_{i},v_{j})}. It follows that ifv andw are the column vectors of the coordinates of two vectorsv andw, one has

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

wherevT{\displaystyle \mathbf {v} ^{\mathsf {T}}} denotes thetranspose of the matrixv.

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

PTBP.{\displaystyle P^{\mathsf {T}}\mathbf {B} P.}

Asymmetric bilinear form is a bilinear formB such thatB(v,w)=B(w,v){\displaystyle B(v,w)=B(w,v)} for everyv andw inV. It follows that the matrix ofB on 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,

(PTBP)T=PTBTP,{\displaystyle (P^{\mathsf {T}}\mathbf {B} P)^{\mathsf {T}}=P^{\mathsf {T}}\mathbf {B} ^{\mathsf {T}}P,}

and the two members of this equation equalPTBP{\displaystyle P^{\mathsf {T}}\mathbf {B} P} if the matrixB is symmetric.

If thecharacteristic of the ground fieldF is 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 fieldR{\displaystyle \mathbb {R} } of thereal numbers, these nonzero entries can be chosen to be either1 or–1.Sylvester's law of inertia is a theorem that asserts that the numbers of1 and of–1 depends only on the bilinear form, and not of the change of basis.

Symmetric bilinear forms over the reals are often encountered ingeometry andphysics, typically in the study ofquadrics and of theinertia of arigid body. In these cases,orthonormal bases are specially useful; this means that one generally prefer to restrict changes of basis to those that have anorthogonal change-of-base matrix, that is, a matrix such thatPT=P1.{\displaystyle 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 theorem asserts 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 theeigenvalues of 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

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Notes

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  1. ^Although a basis is generally defined as a set of vectors (for example, as a spanning set that is linearly independent), thetuple notation is convenient here, since the indexing by the first positive integers makes the basis anordered basis.

References

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  1. ^Anton (1987, pp. 221–237)
  2. ^Beauregard & Fraleigh (1973, pp. 240–243)
  3. ^Nering (1970, pp. 50–52)

Bibliography

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External links

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Basic concepts
Three dimensional Euclidean space
Matrices
Bilinear
Multilinear algebra
Vector space constructions
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