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 25, 2008
A visit to Diagon Alley
Example
Suppose a linear operator L on R 3is represented
in a basis

2 0 0
[u1 , u2 , u3 ] by the diagonal matrix 0 3 0.
0 0 5
This means that Lu1 = 2u1 and Lu2 = 3u2 and Lu3 = 5u3 .
In other words, the basis vectors u1 , u2 , and u3 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.
2
4
Diagonalize the matrix A =
. In other words, find an
−1 −3
invertible matrix S and a diagonal matrix D such that
S −1 AS = D or, equivalently, A = SDS −1 .
Solution. First find theeigenvalues
and eigenvectors of A.
4
Yesterday we saw that
is an eigenvector of A with
−1
1
eigenvalue 1, and
is an eigenvector with eigenvalue −2.
−1
4
1
The matrix S =
is the transition matrix from the
−1 −1
−1
eigenvector basis to the
standard
basis, and the matrix S AS
1
0
is the diagonal matrix
.
0 −2
Continuation
2
4
If A =
, find the power A1000 .
−1 −3
1
0
−1
Solution. Since S AS = D =
, and
0 −2
1
0
D 1000 =
, we have A1000 = SD 1000 S −1 =
0 21000
1
1
0
−1 −1
4
1
−
=
1
4
−1 −1
0 21000
3
1
−4 + 21000 −4 + 4 × 21000
−
.
1 − 21000
1 − 4 × 21000
3
More. Since the exponential function is given by a power series
1 3
1 n
1 2
x + 3!
x + · · · + n!
x + · · ·), we can write
(ex = 1 + x + 2!
1
e
0
1 2
A
D
−1
e := I + A + 2! A + · · · = Se S = S
S −1
0 e−2
1
−4e + e−2 −4e + 4e−2
= −
.
e − e−2
e − 4e−2
3
Application to differential equations
We have
two ways
to solve the system of differential equations
2
4
y0 =
y.
−1 −3
(a) From yesterday,
writethe general solution as
we can 4
1
y(t) = c1 et
+ c2 e−2t
.
−1
−1
(b) With a different
choice ofc1 and
c2 , we can write
c
c
1
1
= SetD S −1
=
y(t) = etA
c2
c2
1
−4et + e−2t −4et + 4e−2t
c1
−
.
t
−2t
t
−2t
c2
e −e
e − 4e
3
c1
y1 (0)
In the second form,
=
.
c2
y2 (0)