Highlights
Math 304
Linear Algebra
From last time:
I
Harold P. Boas
application of eigenvectors to systems of differential
equations
Today:
boas@tamu.edu
I
diagonalization of matrices and applications
June 28, 2006
When is a matrix diagonal?
Example: exercise 1(b), p. 340
3 is represented in a basis
Suppose a linear operator
L on R 
2 0 0
[u1 , u2 , u3 ] by the matrix 0 3 0.
0 0 5
This means that Lu1 = 2u1 and Lu2 = 3u2 and Lu3 = 5u3 .
In other words, the basis vectors are eigenvectors of the
operator L.
A square matrix A is diagonalizable if the linear transformation
x 7→ Ax can be represented in some basis by a diagonal matrix;
in other words, if there is a basis consisting of eigenvectors
of A; equivalently, if there is an invertible matrix S such that
S −1 AS is a diagonal matrix.
5
6
Diagonalize the matrix A =
. In other words, find an
−2 −2
invertible matrix S and a diagonal matrix D such that
S −1 AS = D or A = SDS −1 .
Solution.
First
find the eigenvalues and eigenvectors of A. The
3
vector
is an eigenvector with eigenvalue 1, and the
−2
2
vector
is an eigenvector with eigenvalue 2. The matrix
−1
3
2
S=
is the transition matrix from the eigenvector
−2 −1
basis to the standard
basis,
and the matrix S −1 AS is the
1 0
diagonal matrix
.
0 2
Application to differential equations
Continuation
If A =
5
6
, find A10 .
−2 −2
1 0
1 0
10
Solution. Since
=D=
, and D =
,
0 2
0 210
we have
1 0
3
2
−1 −2
A10 = SD 10 S −1 =
=
−2 −1
2
3
0 210
−3 + 212 −6 + 3 × 211
.
2 − 211
4 − 3 × 210
S −1 AS
More. Since the exponential function is given by a power series
1 2
1 3
1 n
(ex = 1 + x + 2!
x + 3!
x + · · · + n!
x + · · ·), we also have
1 2
A
D
−1
e := I + A + 2! A + · · · = Se S =
e 0
3
2
−1 −2
=
−2 −1
2
3
0 e2
−3e + 4e2 −6e + 6e2
.
2e − 2e2
4e − 3e2
We have
two ways
to solve the system of differential equations
5
6
y0 =
y.
−2 −2
From yesterday,
we can write
the
general solution as
3
2
y(t) = c1 et
+ c2 e2t
.
−2
−1
Alternatively,
solution as
we
can write the
general
c
c
1
1
y(t) = etA
= SetD S −1
=
c2
c2
−3et + 4e2t −6et + 6e2t
c1
.
t
2t
t
2t
c
2e − 2e
4e − 3e
2 c1
y1 (0)
In the second form,
=
.
c2
y2 (0)