Loading [MathJax]/jax/output/HTML-CSS/jax.js
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

10.2: Change of Basis Transformation

( \newcommand{\kernel}{\mathrm{null}\,}\)

Recall that we can associate a matrix AFn×n to every operator TL(V,V). More precisely, the jth column of the matrix A=M(T) with respect to a basis e=(e1,,en) is obtained by expanding Tej in terms of the basis e. If the basis e is orthonormal, then the coefficient of ei is just the inner product of the vector with ei.
Hence,

M(T)=(Tej,ei)1i,jn,
where i is the row index and j is the column index of the matrix.

Conversely, if AFn×n is a matrix, then we can associate a linear operator TL(V,V) to A by setting
Tv=nj=1v,ejTej=nj=1ni=1Tej,eiv,ejei=ni=1(nj=1aijv,ej)ei=i=1(A[v]e)iei,

where (A[v]e)i denotes the ith component of the column vector A[v]e. With this construction, we have M(T)=A.
The coefficients of Tv in the basis (e1,,en) are recorded by the column vector obtained by multiplying the n×n matrix A with the n×1 column vector [v]e whose components ([v]e)j=v,ej.

Example 10.2.1. Given

A=[1ii1],

we can define TL(V,V) with respect to the canonical basis as follows:

T[z1z2]=[1ii1][z1z2]=[z1iz2iz1+z2].

Suppose that we want to use another orthonormal basis f=(f1,,fn) for V. Then, as before, we have v=ni=1v,fifi. Comparing this with v=nj=1v,ejej, we find that

v=ni,j=1v,ejej,fifi=ni=1(nj=1ej,fiv,ej)fi.

Hence,

[v]f=S[v]e,

where

S=(sij)ni,j=1with sij=ej,fi.

The jth column of S is given by the coefficients of the expansion of ej in terms of the basis f=(f1,,fn). The matrix S describes a linear map in L(Fn), which is called the change of basis transformation.

We may also interchange the role of bases e and f. In this case, we obtain the
matrix R=(rij)ni,j=1, where
rij=fj,ei.

Then, by the uniqueness of the expansion in a basis, we obtain
[v]e=R[v]f
so that
RS[v]e=[v]e,for all vV.

Since this equation is true for all [v]eFn, it follows that either RS=I or R=S1. In particular, S and R are invertible. We can also check this explicitly by using the properties of orthonormal bases. Namely,
(RS)ij=nk=1rikskj=nk=1fk,eiej,fk=nk=1ej,fk¯ei,fk=[ej]f,[ei]fFn=δij.

Matrix S (and similarly also R) has the interesting property that its columns are orthonormal to one another. This follows from the fact that the columns are the coordinates of orthonormal vectors with respect to another orthonormal basis. A similar statement holds for the rows of S (and similarly also R).

Example 10.2.2. Let V=C2, and choose the orthonormal bases e=(e1,e2) and f=(f1,f2) with
e1=[10],e2=[01],f1=12[11],f2=12[11].
Then
S=[e1,f1e2,f1e1,f2e2,f2]=12[1111]
and
R=[f1,e1f2,e1f1,e2f2,e2]=12[1111].
One can then check explicitly that indeed
RS=12[1111][1111]=[1001]=I.

So far we have only discussed how the coordinate vector of a given vector vV changes under the change of basis from e to f. The next question we can ask is how the matrix M(T) of an operator TL(V) changes if we change the basis. Let A be the matrix of T with respect to the basis e=(e1,,en), and let B be the matrix for T with respect to the basis f=(f1,,fn). How do we determine B from A? Note that

[Tv]e=A[v]e
so that
[Tv]f=S[Tv]e=SA[v]e=SAR[v]f=SAS1[v]f.

This implies that
B=SAS1.

Example 10.2.3. Continuing Example 10.2.2, let
A=[1111]
be the matrix of a linear operator with respect to the basis e. Then the matrix B with respect to the basis f is given by
B=SAS1=12[1111][1111][1111]=12[1111][2020]=[2000].


This page titled 10.2: Change of Basis Transformation is shared under a not declared license and was authored, remixed, and/or curated by Isaiah Lankham, Bruno Nachtergaele, & Anne Schilling.

Support Center

How can we help?