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Unit 9: Representation of Transformations by Matrices
We motivated the definition (4) of matrix multiplication via operations on the rows of a matrix. Notes
One sees here that a very strong motivation for the definition is to be found in composing linear
transformations. Let us summarize formally.
Theorem 3: Let V, W, and Z be finite-dimensional vector spaces over the field F; let T be a linear
transformation from V into W and U a linear transformation from W into Z. If , ' and '' are
ordered basis for the spaces V, W and Z respectively, if A is the matrix of T relative to the pair
, ' and ''is the matrix of U relative to the pair ', '', then the matrix of the composition UT
relative to the pair , '' is the product matrix C = BA.
We remark that Theorem 3 gives a proof that matrix multiplication is associative – a proof
which requires no calculations.
It is important to note that if T and U are linear operators on a space V and we are representing
by a single ordered basis , then Theorem 3 assumes the simple form UT U T . Thus in
this case, the correspondence which determines between operators and matrices is not only a
vector space isomorphism but also preserve products. A simple consequence of this is that the
linear operator T is invertible if and only if T is an invertible matrix. For, the identity operator
I is represented by the identity matrix in any order basis, and thus
UT = TU = I
is equivalent to
U T T U . I
Of course, when T is invertible
T 1 T 1 .
Now we should like to inquire what happens to representing matrices when the ordered basis
is changed. For the sake of simplicity, we shall consider this question only for linear operators
on a space V, so that we can use a single ordered basis. The specific question is this. Let T be a
linear operator on the finite-dimensional space V, and let
= 1 ,..., n and '= ' 1 ,..., ' n
be two ordered basis for V. How are the matrices T and T ' ' related? There is a unique
(invertible) n×n matrix P such that
P ...(5)
'
for every vector in V. It is the matrix P P ,...,P where Pj ' . By definition
1 n j
T T . ...(6)
Applying (5) to the vector T ,we have
T P T . ...(7)
'
Combining (5), (6) and (7), we obtain
T P ' P T '
P 1 T P T
' '
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