Pengguna:Kekavigi/bak pasir

Ruang vektor pada yang dibangun dari basis berisi vektor-vektor berwarna ungu, dapat dibangun pula dari basis berisi vektor-vektor berwarna merah. Dengan menyusun setiap vektor berwarna ungu menjadi kombinasi linear vektor-vektor berwarna merah (dan sebaliknya), pernyataan yang disampaikan menggunakan basis ungu juga dapat disampaikan menggunakan basis merah (dan sebaliknya).

Dalam aljabar linear, basis terurut memungkinkan setiap elemen pada sebarang ruang vektor berdimensi $n$ dinyatakan dalam bentuk vektor koordinat, yakni suatu barisan $n$-skalar yang disebut [koordinat]. Untuk dua basis berbeda dari suatu ruang vektor, vektor koordinat yang menyatakan elemen $v$ atas basis pertama, umumnya berbeda dengan vektor koordinat yang menyatakan elemen yang sama, namun atas basis yang kedua. Perubahan basis adalah tindakan mengubah penyataan-pernyataan matematika pada suatu basis, ke pernyataan-pernyataan yang sepadan pada basis yang lain.

Perubahan basis tersebut dihasilkan dari rumus perubahan-basis yang menyatakan koordinat-koordinat relatif pada satu basis dalam bentuk koordinat-koordinat relatif basis yang lainnya. Menggunakan matriks, rumus ini dapat dituliskan sebagai

$\mathbf {x} _{\mathrm {lama} }=\mathbf {A} \,\mathbf {x} _{\mathrm {baru} },$

dengan "lama" dan "baru" masing-masing merujuk pada basis lama dan basis baru, $\mathbf {x} _{\mathrm {lama} }$ dan $\mathbf {x} _{\mathrm {baru} }$ adalah vektor kolom dari koordinat vektor yang sama menurut kedua basis, dan $\mathbf {A}$ adalah matriks perubahan-basis (juga disebut matriks transisi) yang kolom-kolomnya menyatakan koordinat vektor-vektor basis lama di basis baru.

Artikel ini fokus membahas ruang vektor dimensi hingga. Akan tetapi, banyak prinsip yang disampaikan disini juga berlaku pada ruang vektor dimensi tak-hingga.

Contoh

Misalkan kita ingin menentukan vektor koordinat pada ruang vektor Euklides $\mathbb {R} ^{2},$ yang dihasilkan dari rotasi sebesar $t.$ Ruang vektor sebelum dirotasi memiliki basis standar $v_{1}=(1,0)$ dan $v_{2}=(0,1);$ sebut ini sebagai basis "lama". Setelah ruang vektor dirotasi, basis tersebut ikut berotasi, dan berubah menjadi $w_{1}=(\cos t,\sin t)$ dan $w_{2}=(-\sin t,\cos t);$ sebut ini sebagai basis "baru". Lalu, matriks perubahan basis akibat rotasi dapat dituliskan sebagai

$${\begin{bmatrix}\cos t&-\sin t\\\sin t&\cos t\end{bmatrix}}.$$

Rumus perubahan basis menyatakan bahwa, jika ${\textstyle (y_{1},\,y_{2})}$ adalah koordinat baru hasil rotasi dari vektor koordinat $(x_{1},\,x_{2}),$ maka

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

Sehingga,

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

Hubungan tersebut dapat dibuktikan dengan menunjukkan $(x_{1},\,x_{2})$ dan ${\textstyle (y_{1},\,y_{2})}$ merujuk pada objek yang sama,

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

Rumus perubahan basis

Misalkan $B_{\mathrm {lama} }=(v_{1},\,\ldots ,\,v_{n})$ adalah basis dari suatu ruang vektor dimensi hingga V atas suatu lapangan F. Seorang dapat mendefinisikan $n$ vektor baru $w_{j};$ (dengan $j=1,\,\dots ,\,n$) berdasarkan koordinat vektor tersebut $a_{i,j}$ atas $B_{\mathrm {lama} }\colon$

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

Misalkan

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

sebagai matriks yang kolom ke-j-nya dibentuk dari koordinat $w_{j}$. (Di bagian ini dan seterusnya, indeks i selalu digunakan untuk merujuk baris di $\mathbf {A}$ dan $v_{i},$ sedangkan indeks j selalu digunakan untuk merujuk kolom di $\mathbf {A}$ dan $w_{j};$ konvensi ini berguna untuk menghindari kesalahan dalam perhitungan.)

Barisan terurut $B_{\mathrm {baru} }=(w_{1},\ldots ,w_{n})$ disebut basis dari V jika dan hanya jika matriks $\mathbf {A}$ terbalikkan, atau setara dengan itu, jika determinannya tidak bernilai nol. Dalam kasus ini, $\mathbf {A}$ disebut matriks perubahan basis dari basis $B_{\mathrm {lama} }$ ke basis $B_{\mathrm {baru} }$.

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Lebih lanjut, untuk sebarang vektor $z\in V,$ misalkan $(x_{1},\ldots ,x_{n})$ adalah koordinat $z$ atas $B_{\mathrm {lama} }$, dan $(y_{1},\ldots ,y_{n})$ adalah koordinatnya atas $B_{\mathrm {baru} }$, kita memiliki hubungan

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

The change-of-basis formula expresses the coordinates over the old basis in term 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 of z over $B_{\mathrm {old} }$ 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.

In terms of linear maps

Normally, a 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 bases of implied vector spaces, and to the fact that the choice of a basis induces an isomorphism between a vector space and Fⁿ, where 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 F be a field, the set $F^{n}$ of the n-tuples is a F-vector space whose addition and scalar multiplication are defined component-wise. Its standard basis is the basis that has as its ith element the tuple with all components equal to 0 except the ith that is 1.

A basis $B=(v_{1},\ldots ,v_{n})$ of a F-vector space V defines a linear 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 matrix A, one could consider it the matrix of an endomorphism $\psi _{A}$ of $F^{n}.$ Finally, define

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

(where $\circ$ denotes function 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

A function that has a vector space as its domain is commonly specified as a multivariate function whose variables are the coordinates on some basis of the vector on which the function is applied.

When the basis is changed, the expression 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, if f(x) is the expression of the function in terms of the old coordinates, and if x = Ay is the change-of-base formula, then f(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 no matrix inversion is needed here.

As the change-of-basis formula involves only linear 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,
• a polynomial function,
• a continuous function,
• a differentiable function,
• a smooth function,
• an analytic 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 of manifolds, as this allows extending the concepts of continuous, differentiable, smooth and analytic functions to functions that are defined on a manifold.

Linear maps

Consider a linear map T: W → V from a vector space W of dimension n to a vector space V of dimension m. It is represented on "old" bases of V and W by a m×n matrix M. A change of bases is defined by an m×m change-of-basis matrix P for V, and an n×n change-of-basis matrix Q for W.

On the "new" bases, the matrix of T is

$P^{-1}MQ.$

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

Endomorphisms

Endomorphisms, are linear maps from a vector space V 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, if M is the square matrix of an endomorphism of V over an "old" basis, and P is a change-of-basis matrix, then the matrix of the endomorphism on the "new" basis is

$P^{-1}MP.$

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

Bilinear forms

A bilinear form on a vector space V over a field F is a function V × V → F which is linear in both arguments. That is, B : V × V → F is bilinear if the maps $v\mapsto B(v,w)$ and $v\mapsto B(w,v)$ are linear for every fixed $w\in V.$

The matrix B of a bilinear form B on a basis $(v_{1},\ldots ,v_{n})$ (the "old" basis in what follows) is the matrix whose entry of the ith row and jth column is B(i, j). It follows that if v and w are the column vectors of the coordinates of two vectors v and w, one has

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

where $\mathbf {v} ^{\mathsf {T}}$ denotes the transpose of the matrix v.

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

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

A symmetric bilinear form is a bilinear form B such that $B(v,w)=B(w,v)$ for every v and w in V. It follows that the matrix of B on any basis is symmetric. 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 matrix B is symmetric.

If the characteristic of the ground field F is not two, then for every symmetric bilinear form there is a basis for which the matrix is diagonal. 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 the real numbers, these nonzero entries can be chosen to be either 1 or –1. Sylvester's law of inertia is a theorem that asserts that the numbers of 1 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 in geometry and physics, typically in the study of quadrics and of the inertia of a rigid body. In these cases, orthonormal bases are specially useful; this means that one generally prefer to restrict changes of basis to those that have an orthogonal change-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. The Spectral 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 the eigenvalues of the initial matrix on the diagonal. It follows that, over the reals, if the matrix of an endomorphism is symmetric, then it is diagonalizable.

See also

• Active and passive transformation
• Covariance and contravariance of vectors
• Integral transform, the continuous analogue of change of basis.

Bibliography

• Anton, Howard (1987), Elementary Linear Algebra (edisi ke-5th), New York: Wiley, ISBN 0-471-84819-0 
• 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, ISBN 0-395-14017-X 
• Nering, Evar D. (1970), Linear Algebra and Matrix Theory (edisi ke-2nd), New York: Wiley, LCCN 76091646 

External links

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