matrix representation of a linear transformation (original) (raw)

Linear transformations as matrices

Let V,W be vector spaces (over a common field k) of dimension n and m respectively. Fix bases A={v1,…,vn} and B={w1,…,wm} for V and W respectively. We shall order these bases so that vi<vj and wi<wj whenever i<j. To distinguish an ordinary set from an ordered set, we shall adopt the notation ⟨v1,…,vn⟩ to mean the set {v1,…,vn} with ordering vi≤vj whenever i≤j. The importance of ordering these bases will be apparent shortly.

For any linear transformation T:V→W, we can write

for each j∈{1,…,n} and αi⁢j∈k. We define the matrix associated with the linear transformation T and ordered bases A,B by

where 1≤i≤n and 1≤j≤m. [T]BA is a m×n matrix whose entries are in k. When A=B, we often write [T]A:=[T]AA. In addition, when both ordered bases are standard bases En,Em ordered in the obvious way, we write [T]:=[T]EmEn.

Examples.

1. 1. Let T:ℝ3→ℝ4 be given by

      T⁢(xyz)=(x+2⁢y+zz-x+y-5⁢z3⁢x+2⁢z).
      Using the standard ordered bases
      E3=⟨(100),(010),(001)⟩⁢ for ⁢ℝ3 and E4=⟨(1000),(0100),(0010),(0001)⟩⁢ for ⁢ℝ4
      --------------------------------------------------------------------------------
      ordered in the obvious way. Then,
      T⁢(100)=(10-13),T⁢(010)=(2010),T⁢(001)=(11-52),
      ----------------------------------------------------
      so the matrix [T]E4E3 associated with T and the standard ordered bases E3 and E4 is the 4×3 matrix

2. 1. Let T be the same linear transformation as above. However, let E3′ be the same basis as E3 except that the order is reversed: e3<e2<e1. Then

      [T]E4E3′=(121100-51-1203).
      Note that this matrix is just the matrix from the previous example except that the first and the last columns have been switched.

3. 1. Again, let T be the same as before. Now, let E4′ be the ordered basis whose elements are those of E4 but the order is now given by e2<e1<e4<e3. Then

      [T]E4′E3′=(100121203-51-1).
      Note that this matrix is just the matrix from the previous example except that the first two rows and the last two rows have been interchanged.

Remarks.

• •
  From the examples above, we note several important features of a matrix representation of a linear transformation:
  1. (a)
     the matrix depends on the bases given to the vector spaces
  2. (b)
     the ordering of a basis is important
  3. (c)
     switching the order of a given basis amounts to switching columns and rows of the matrix, essentially multiplying a matrix by a permutation matrix.
• •
  Some basic properties of matrix representations of linear transformations are
  1. (a)
     If T:V→W is a linear transformation, then [r⁢T]BA=r⁢[T]BA, where A,B are ordered bases for V,W respectively.
  2. (b)
     If S,T:V→W are linear transformations, then [S+T]BA=[S]BA+[T]BA, where A and B are ordered bases for V and W respectively.
  3. (c)
     If S:U→V and T:V→W, then [T⁢S]CA=[T]CB⁢[S]BA, where A,B,C are ordered bases for U,V,W respectively.
  4. (d)
• •
  We could have represented all vectors as row vectors. However, doing so would mean that the matrix representation M1 of a linear transformation T would be the transpose of the matrix representation M2 of T if the vectors were represented as column vectors: M1=M2T, and that the application of the matrices to vectors would be from the right of the vectors:

  (abc)⁢(10-13201011-52) instead of (121001-11-5302)⁢(abc).

Matrices as linear transformations

Every m×n matrix A over a field k can be thought of as a linear transformation from kn to km if we view each vector v∈kn as a n×1 matrix (a column) and the mapping is done by the matrix multiplication A⁢v, which is a m×1 matrix (a column vector in km). Specifically, we define TA:kn→km by

It is easy to see that TA is indeed a linear transformation. Furthermore, [TA]=[TA]EmEn=A, since the representation of vectors as n-tuples of elements in k is the same as expressing each vector under the standard basis (ordered) in the vector space kn. Below we list some of the basic properties:

1. 1. Tr⁢A=r⁢TA, for any r∈k,
2. 1. TA+TB=TA+B, where A,B are m×n matrices over k
3. 1. TA∘TB=TA⁢B, where A is an m×n matrix and B is an n×p matrix over k
4. 1. TA is invertible iff A is an invertible matrix.

Remark. As we can see from the discussion above, if we fix sets of base elements for a vector space V and W, there is a one-to-one correspondence between the set of matrices (of the same size) over the underlying field k and the set of linear transformations from V to W.

References

• 1 Friedberg, Insell, Spence. Linear Algebra. Prentice-Hall Inc., 1997.