Download Applying transformations in succession Suppose that A and B are 2

Applying transformations in succession
Suppose that A and B are 2 × 2 matrices representing the maps TA
and TB . Then,
TB (TA(x)) = B (TA(x)) = B (A x) = (B A) x,
so the combined transformation is represented by the matrix B A.
That is,
TB ◦ TA = TBA.
Note that the above is “first TA, then TB ”, and not the other way
around!
Inverses of transformations
The matrix for a rotation or a reflection is invertible: their determinants
are non-zero. Suppose that A is the 2 × 2 matrix representing TA,
and it has inverse A−1. Then
TA−1 (TA (x)) = A−1(Ax) = (A−1A)x = I2x = x,
and similarly, TA(TA−1 (x)) = x. So to get the matrix for the inverse,
we just take the inverse of the matrix!
Eigenvalues
Definition A number λ is called an eigenvalue for A if there is a
non-zero n × 1 vector v such that
A v = λ v.
The vector v is called an eigenvector for λ.
Example
Let A =
2 1
−4 7
!
. Then
A
1
1
!
=
3
3
!
=3
1
1
!
,
so λ = 3 is an eigenvalue of A with (1, 1)T a corresponding eigenvector.
How can we find all of the eigenvectors of a given square matrix?
Theorem Let A be a 2 × 2 matrix. Then λ is an eigenvalue of A if
and only if (A − λ I2) is singular if and only if det (A − λ I2) = 0.
Example
Find all eigenvalues of the 2 × 2 matrix A =
det (A − λ I2) = det
= det
5 −6
2 −2
!
1 0
0 1
−λ
−
:
λ 0
0 λ
!
!!
!!
5−λ
−6
2
−2 − λ
(5 − λ) (−2 − λ) − (−6)2
−10 − 5 λ + 2 λ + λ2 + 12
λ2 − 3 λ + 2
(λ − 1) (λ − 2).
= det
=
=
=
=
5 −6
2 −2
!
5 −6
2 −2
!
Now, this determinant is zero exactly when λ = 1 or 2, so these are
the eigenvalues of A.
General method for the eigenvalue problem
Definition Let A be an n × n matrix. Then the characteristic polynomial p for A is
p(λ) = |A − λ In|.
Corollary. The eigenvalues of A are the roots of the characteristic
polynomial p. That is, the numbers λ such that p(λ) = 0.
Solving the eigenvalue problem
1. Use determinants to calculate p(λ) = |A − λ In |.
2. Find the values λ such that p(λ) = 0.
3. For each λ, use linear algebra to solve the equation (A −
λ In) v = 0.
Example Find the eigenvectors in the previous example. First let
λ = 1. Then
A − λ In =
5 −6
2 −2
!
−
1 0
0 1
!
=
4 −6
2 −3
!
.
We need to solve
4 −6
2 −3
!
0
0
u=
!
.
The augmented matrix is
4 −6 0
2 −3 0
!
1 R , R → 1 R give the RREF
and the row operations R2 → R2 − 2
1
1
4 1
3
1 −2
0
0
0 0
!
.
The general solution to this system is u2 = t, u1 = 3
2 t. Taking t = 2
(any non-zero t would do), we obtain an eigenvector
u=
3
2
!
.
It is easy to check that
Au =
5 −6
2 −2
!
3
2
!
=
3
2
!
= 1 u,
so u is indeed an eigenvector for the eigenvalue 1. A similar calculation shows that v = (2, 1)T is an eigenvector for λ = 2.