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Section4.7Invertible matrices and Coordinate Systems

Objectives
  1. Learn to use invertible matrices to convert between coordinate systems.
  2. Learn to represent linear transformations with respect to given bases.
  3. Recipes: compute the (B,C) -matrix of a linear transformation.
  4. Vocabulary: (B,C) -matrix.

In this section, we study matrix representations for linear transformations with respect to bases for the domain and the codomain. We'll also get another perspective on bases in terms of the linear transformations they represent, and we'll see how to convert between different coordinate systems using matrices and their inverses.

Indeed, the columns of B span V by definition, and they're linearly independent if and only if T is one-to-one, using theorem in Section 4.2.

In particular, a list B=(v1,...,vn) of vectors in Rn forms a basis for all of Rn if and only if the square n×n matrix B with columns v1,...,vn is invertible. These columns are clearly the coordinate vectors of the vi with respect to the standard basis E=(e1,...,en) for Rn. Conversely, the columns of the inverse matrix B−1 are the B -coordinates of the standard basis vectors:

B−1=C|||B[e1]B[e2]···B[en]|||D.

This is because BB−1=In, so BB[ei]=ei, which is the definition of B[ei] being the B -coordinates of ei. Multiplying by B−1 and taking linear combinations we get:

If B is an invertible n×n square matrix, whose columns thus form a basis B of Rn, the B -coordinates of any vector x∈Rn is given by

B[x]=B−1xandhenceBB[x]=x.

This says that (multiplication by) B changes from the B -coordinates to the usual coordinates, and B−1 changes from the usual coordinates to the B -coordinates.

Definition

Suppose we're given bases B=(v1,...,vn) for Rn and C=(w1,...,wm) for Rm. Let T:Rn→Rm be a linear transformation. The (B,C) -matrix of T is the m×n matrix

C[T]B=C|||C[T(v1)]C[T(v2)]···C[T(vn)]|||D.

This generalizes the definition of the standard matrix of T: If we let B=En and C=Em be the standard bases for Rn and Rm, then Em[T]En=[T], the standard matrix of T.

By the discussion above, if E is the standard basis for Rn and B=(v1,...,vn) is another basis for Rn, so the square n×n matrix B with columns v1,...,vn is invertible, then

B=E[IdRn]BandB−1=B[IdRn]E.

If B is an invertible n×n square matrix, whose columns thus form a basis B of Rn, and C is an invertible m×m square matrix, whose columns thus form a basis C of Rm, then the (B,C) -matrix of a linear transformation T:Rn→Rm is given by the matrix product

C[T]B=C−1[T]B,

where [T] is the standard matrix for T.

One way to read this formula is right-to-left: The multiplying by B converts from /cB -coordinates to standard coordinates in Rn; multiplying by the standard matrix [T] then results in the standard coordinates in Rm of the result of acting with the transformation T, while finally multiplying by C−1 converts to the C -coordinates. Altogether, the matrix product C−1[T]B takes B -coordinates of a vector x∈Rn to the C -coordinates of the image T(x)∈Rm

Conversely, we can recover the standard matrix of T as

[T]=CC[T]BB−1;

this says that to do T in standard coordinates is the same as to first convert to B -coordinates, then do T in the (B,C) -coordinates, and finally convert from C -coordinates back to standard coordinates.

Recipe: Computing T(x) in terms of the (B,C) -matrix

Suppose that A=[T]=CDB−1 is the standard matrix for T:Rn→Rm, where B is the invertible matrix corresponding to the basis B for Rn, C is the invertible matrix corresponding to the basis C for Rm, and D=C[T]B is the (B,C) -matrix for T. To compute T(x)=Ax, for some x∈Rn, one can do the following:

  1. Multiply x by B−1, which changes to the B -coordinates: B[x]=B−1x.
  2. Multiply this by D: DB[x]=DB−1x.
  3. Interpreting this vector as a C -coordinate vector, we multiply it by C to change back to the usual coordinates for Rm: Ax=CDB−1x=CDB[x].

To summarize: if A=CDB−1, then A and D do the same thing, only in different coordinate systems for Rn and Rm.

Example

Consider the matrices

A=K0101LB=K0−110LC=K111−1LD=K1000L.

One can verify that A=CDB−1: try it yourself. Let v1=A01B and v2=A−10B, the columns of B corresponding to the basis B=(v1,v2) of R2. Let w1=A11B and w2=A1−1B, the columns of C corresponding to the basis C=(w1,w2), another basis of R2.

The matrix D is diagonal with a one and a zero: it keeps the x -direction and zeroes out the y -direction.

To compute Ax, first we multiply by B−1 to find the B -coordinates of x, then we multiply by D, then we multiply by C. For instance, let x=A12B.

  1. We see from the B -coordinate grid below that x=2v1−v2. Therefore, B−1x=B[x]=A2−1B.
  2. Multiplying by D zeroes out the y -coordinate: DB[x]=A20B.
  3. Interpreting A20B as a C -coordinate vector, we multiply by C to get
    Ax=CK20L=2w1=K22L.
    Of course, this vector lies at (2,0) on the C -coordinate grid.

B-coordinatesB[x]usualEn-coordinates,n=2xC-coordinatesDB[x]usualEm-coordinates,m=2Ax=CDB−1xmultiplybyB−1multiplybyCmultiplybyDmultiplybyA

To summarize:

  • D retains the e1 -direction and zeroes out the e2 -direction.
  • A maps v1 -direction onto the w1 -direction and maps the v2 -direction to zero.
Proof
Proof

This is a very simple kind of diagonalisation theorem, which is made possible by the fact that we can choose bases for the domain and the codomain independently. In Chapter 6 we'll see that the situation is much more subtle for linear transformations T:Rn→Rn where the domain and the codomain coincide and we want to use the same basis for both.