Change of basis

Example

Consider the Euclidean vector space ${\displaystyle {\mathbb{R}}^2}$ and its standard basis, consisting of the vectors ${\displaystyle v_1}$ and ${\displaystyle v_2}$ with column vectors

${\displaystyle {\mathbf{v}}_{\mathbf{1}}=\left[\begin{array}{c}1\\ 0\end{array}\right]}$ and ${\displaystyle {\mathbf{v}}_{\mathbf{2}}=\left[\begin{array}{c}0\\ 1\end{array}\right].}$

Rotating them by an angle of ${\displaystyle t}$ gives a "new" basis, formed by the vectors ${\displaystyle w_1}$ and ${\displaystyle w_2}$ with column vectors

${\displaystyle {\mathbf{w}}_{\mathbf{1}}=\left[\begin{array}{c}\cos t\\ \sin t\end{array}\right]}$ and ${\displaystyle {\mathbf{w}}_{\mathbf{2}}=\left[\begin{array}{c}-\sin t\\ \cos t\end{array}\right].}$

So, the change-of-basis matrix is

${\displaystyle \left[\begin{array}{cc}\cos t& -\sin t\\ \sin t& \cos t\end{array}\right].}$

The change-of-basis formula asserts that, for a vector with "old" coordinates ${\displaystyle (x_1,x_2)}$ and "new" coordinates ${\displaystyle (y_1,y_2),}$ one has

${\displaystyle \left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{cc}\cos t& -\sin t\\ \sin t& \cos t\end{array}\right]\left[\begin{array}{c}{y}_1\\ y_2\end{array}\right].}$

That is,

${\displaystyle \{\begin{array}{l}{x}_1=y_1\cos t-y_2\sin t\\ x_2=y_1\sin t+y_2\cos t.\end{array}}$

This may be shown by writing

${\displaystyle \begin{array}{cc}{y}_1{w}_1+y_2{w}_2& =y_1((\cos t)v_1+(\sin t)v_2)+y_2(-(\sin t)v_1+(\cos t)v_2)\\ & =(y_1\cos t-y_2\sin t)v_1+(y_1\sin t+y_2\cos t)v_2\\ & =x_1{v}_1+x_2{v}_2.\end{array}}$

In terms of linear maps

Usually, 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 a linear isomorphism between a vector space and ${\displaystyle F^n}$, where F is the ground field of scalars. When only one basis is considered for each vector space, it is convenient 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 ${\displaystyle F^n}$ of the n-tuples is an F-vector space whose addition and scalar multiplication are defined component-wise. Its standard basis is the basis that has as its i-th element the tuple with all components equal to 0 except the i-th one, equal to 1.

A basis ${\displaystyle B=(v_1,\dots ,v_n)}$ of an F-vector space V defines a linear isomorphism ${\displaystyle \phi :F^n\to V}$ by

${\displaystyle \phi (x_1,\dots ,x_n)=\sum _{i=1}^n{x}_i{v}_i.}$

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

Let ${\displaystyle B_{\mathrm{o}\mathrm{l}\mathrm{d}}=(v_1,\dots ,v_n)}$ be the "old" basis of a change of basis, and ${\displaystyle {\phi }_{\mathrm{o}\mathrm{l}\mathrm{d}}}$ the associated isomorphism. Given a change-of-basis (invertible) matrix A, one could consider it the matrix of an automorphism (a bijective endomorphism) ${\displaystyle {\psi }_A}$ of ${\displaystyle F^n.}$ Finally, define

${\displaystyle {\phi }_{\mathrm{n}\mathrm{e}\mathrm{w}}={\phi }_{\mathrm{o}\mathrm{l}\mathrm{d}}\circ {\psi }_A}$

(where ${\displaystyle \circ }$ denotes function composition), and

${\displaystyle B_{\mathrm{n}\mathrm{e}\mathrm{w}}={\phi }_{\mathrm{n}\mathrm{e}\mathrm{w}}\left({\phi }_{\mathrm{o}\mathrm{l}\mathrm{d}}^{-1}(B_{\mathrm{o}\mathrm{l}\mathrm{d}})\right).}$

A straightforward verification shows that this definition of ${\displaystyle B_{\mathrm{n}\mathrm{e}\mathrm{w}}}$ is the same as that in the preceding section.

Now, by composing the equation ${\displaystyle {\phi }_{\mathrm{n}\mathrm{e}\mathrm{w}}={\phi }_{\mathrm{o}\mathrm{l}\mathrm{d}}\circ {\psi }_A}$ with ${\displaystyle {\phi }_{\mathrm{o}\mathrm{l}\mathrm{d}}^{-1}}$ on the left and ${\displaystyle {\phi }_{\mathrm{n}\mathrm{e}\mathrm{w}}^{-1}}$ on the right, one gets

${\displaystyle {\phi }_{\mathrm{o}\mathrm{l}\mathrm{d}}^{-1}={\psi }_A\circ {\phi }_{\mathrm{n}\mathrm{e}\mathrm{w}}^{-1}.}$

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

${\displaystyle {\phi }_{\mathrm{o}\mathrm{l}\mathrm{d}}^{-1}(v)={\psi }_A\left({\phi }_{\mathrm{n}\mathrm{e}\mathrm{w}}^{-1}(v)\right),}$

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-basis 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" ones may seem unnatural, but appears as useful, because 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 L: V → W from a vector space V of dimension n to a vector space W of dimension m. It is represented on "old" bases of V and W by an m×n matrix M. A change of basis is defined by an n×n change-of-basis matrix P for V, and an m×m change-of-basis matrix Q for W.

On the "new" bases, the matrix representation of L is

${\displaystyle M^{\prime }=Q^{-1}MP.}$

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 on an "old" basis, and P is a change-of-basis matrix, then the matrix of the endomorphism on the "new" basis is

${\displaystyle M^{\prime }=P^{-1}MP.}$

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 ${\displaystyle \Phi :V\times V\to F}$ which is linear in both arguments. That is, ${\displaystyle \Phi }$ is bilinear if the maps ${\displaystyle v\mapsto \Phi (v,w)}$ and ${\displaystyle v\mapsto \Phi (w,v)}$ are linear for every fixed ${\displaystyle w\in V.}$

The matrix ${\displaystyle \mathbf{\Phi }}$ of a bilinear form ${\displaystyle \Phi }$ on a basis ${\displaystyle (v_1,\dots ,v_n)}$ (the "old" basis) is the matrix whose entry of the i-th row and j-th column is ${\displaystyle \Phi (v_i,v_j)}$. It follows that for two vectors v and w with coordinate column vectors v and w, one has

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

where ${\displaystyle {\mathbf{v}}^{\mathsf{T}}}$ denotes the transpose of the column vector v.

For a change of basis with matrix P, a straightforward computation shows that the matrix of the bilinear form on the "new" basis is

${\displaystyle {\mathbf{\Phi }}^{\prime }=P^{\mathsf{T}}\mathbf{\Phi }P.}$

A symmetric bilinear form is a bilinear form S such that ${\displaystyle S(v,w)=S(w,v)}$ for every v and w in V. It follows that the matrix of S on any basis is symmetric. This implies that the property of being a symmetric matrix must be kept by the above change-of-basis 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. Thus,

${\displaystyle (P^{\mathsf{T}}\mathbf{S}P{)}^{\mathsf{T}}=P^{\mathsf{T}}{\mathbf{S}}^{\mathsf{T}}P;}$

finally, since the matrix S is symmetric,

${\displaystyle (P^{\mathsf{T}}\mathbf{S}P{)}^{\mathsf{T}}=P^{\mathsf{T}}\mathbf{S}P.}$

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 ${\displaystyle \mathbb{R}}$ 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 depend only on the bilinear form, and not on 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 prefers to restrict changes of basis to those that have an orthogonal change-of-basis matrix, that is, a matrix such that ${\displaystyle P^{\mathsf{T}}=P^{-1}.}$ Such matrices have the fundamental property that the change-of-basis 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.