Download Find the standard matrix of the gi

Let’s see some more examples of finding standard matrix of a matrix transformation
Example: Find the standard matrix of the given operators
1. T : R3 −→ R3 , reflection through the xy-plane
2. T : R3 −→ R3 , reflection through the plane x=z
3. T : R3 −→ R3 , Dilation with factor k=2
Solution:
1. .


1 0 0
[T ] = 0 1 0 
0 0 −1
2. .
n = (1, 0, −1), T (u) = u−2projn u= (u1 , u2 , u3 )−(u1 −u3 , 0, u3 −u1 ) = (u3 , u2 , u1 )


0 0 1
[T ] = 0 1 0
1 0 0
1
3. .


k 0 0
[T ] = 0 k 0
0 0 k
Please look at book for more examples on standard matrix of contraction, rotation, reflection, compression operators.
Section 4.10 Properties of Matrix Transformations
Composition of Matrix Transformations:
Let T1 : Rn → Rk and T2 : Rk → Rm be two matrix transformations for which codomain
of T1 = domain of T2 . Then for each x ∈ Rn we can compute T2 (T1 (x)) : Rn → Rm .
Application of T1 followed by T2 produces a transformation from Rn to Rm . This is called
the composition of T2 with T1 and is denoted by T2 ◦ T1 and T2 ◦ T1 (x) = T2 (T1 (x)).
Lets look at the matrix side:
Recall that we denote by [T1 ]k×n standart matrix for T1 , [T2 ]m×k standart matrix for T2 .
T1 (x) = [T1 ]x,
T2 ◦ T1 (x) = T2 (T1 (x)) = [T2 ]T1 (x) = [T2 ][T1 ]x
⇒ [T2 ◦ T1 ] = [T2 ][T1 ]
Example: Let T1 (x1 , x2 ) = (x1 + x2 , x1 − x2 ) and T2 (x1 , x2 ) = (3x1 , 2x1 + 4x2 ).
Find standart matrix for T2 ◦ T1 and T1 ◦ T2 .
1 1
3 0
Solution: [T1 ] =
, [T2 ] =
1 −1
2 4
2
5 4
[T1 ◦ T2 ] = [T1 ][T2 ] =
1 −4
3 3
[T2 ◦ T1 ] = [T2 ][T1 ] =
6 −2
In general, if we have T1 , T2 , . . . , Tn , n matrix transformations with the domains and
codomains match appropriately, then we can define the composition T1 ◦ T2 ◦ . . . ◦ Tn (x)
and the standard matrix of this transformation will be [T1 ][T2 ] . . . [Tn ].
One-to-one Matrix Transformations and Inverse of a one-to-one Matrix
Operator
To be able to define inverse of a matrix transformation we need to have one-to-one matrix
operator
Definition: A matrix transformation T : Rn → Rm is said to be one-to-one if T
maps distinct vectors (points) in Rn into distinct vectors (points) in Rm . That is for each
w in the range of T, there is exactly one vector x such that T (x) = w.
Theorem 4.10.1: If A is an n × n matrix and TA : Rn → Rn is the corresponding
matrix operator, then the following statements are equivalent.
(a) A is invertible
(b) The range of TA is Rn
(c) TA is one-to-one.
This Theorem is implied by equivalent statements (a),(e) and (f) of Theorem 4.8.10.
Definition: If T : Rn → Rn is a one-to-one matrix operator then we can define
T −1 : Rn ⇒ Rn a matrix operator called the inverse of T so that
T ◦ T −1 (w) = T (T −1 (w)) = w
T −1 ◦ T (x) = T −1 (T (x)) = x
Note: If T is a one-to-one matrix operator on Rn , then the standard matrix for T is
invertible and its inverse is the standard matrix for T −1 , that is
[T −1 ] = [T ]−1
Example: Determine whether the matrix operator T defined by the equations is
one-to-one, if so find T −1 .
w1 = x1 − 2x2 + 2x3
w2 = 2x1 + x2 + x3
w3 = x1 + x2
Solution: T : R3 → R3 , (x1 , x2 , x3 ) = (x1 − 2x2 + 2x3 , 2x1 + x2 + x3 , x1 + x2 )


1 −2 2
[T ] = 2 1 1. By Theorem 4.10.1 T is one-to-one if [T] is invertible.
1 1 0
3
Considerdet ([T ]) = −1hence T is invertible and the standart matrix for inverse of T is
1 −2 4
−1

[T ] = −1 2 −3 and
−1 3 −5
T −1 (w1 , w2 , w3 ) = (w1 − 2w2 + 4w3 , −w1 + 2w2 − 3w3 , −w1 + 3w2 − 5w3 )
We will add two more equivalent statements to Theorem 4.8.10:
Theorem 4.10.4 Equivalent statements: If A is an n × n matrix, then the
following statements are equivalent.
(a) A is invertible
(b) Ax=0 has only the trivial solution.
(c) The reduced row echelon form of A is I
(d) A is expressible as a product of elementary matrices.
(e) Ax=b is consistent for every n × 1 matrix b.
(f) Ax=b has exactly one solution for every n × 1 matrix b.
(g) det (A) 6= 0
(h) The column vectors of A are linearly independent.
(i) The row vectors of A are linearly independent.
(j) The column vectors of A span Rn .
(k) The row vectors of A span Rn .
(l) The column vectors of A form a basis for Rn .
(m) The row vectors of A form a basis for Rn .
(n) A has rank n.
(o) A has nullity 0.
(p) The orthogonal complement of the null space of A is Rn .
(q) The orthogonal complement of the row space of A is {0}.
(r) The range of TA is Rn .
(s) TA is one-to-one.
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