Diagonalization of Matrices and Applications in Linear Algebra, Study notes of Linear Algebra

The concept of diagonalization of matrices in linear algebra. It explains when a matrix is diagonal and provides an example of diagonalizing a matrix using eigenvectors and an invertible matrix. The document also introduces the diagonal matrix d and the transition matrix s, and shows how to find the matrix a10 and the exponential function ea. The document concludes with an application of diagonalization to systems of differential equations.

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Pre 2010

Uploaded on 02/10/2009

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Math 304
Linear Algebra
Harold P. Boas
June 28, 2006
Highlights
From last time:
Iapplication of eigenvectors to systems of differential
equations
Today:
Idiagonalization of matrices and applications
When is a matrix diagonal?
Suppose a linear operator Lon R3is represented in a basis
[u1,u2,u3]by the matrix
200
030
005
.
This means that Lu1=2u1and Lu2=3u2and Lu3=5u3.
In other words, the basis vectors are eigenvectors of the
operator L.
A square matrix Ais diagonalizable if the linear transformation
x7→ Axcan 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 Ssuch that
S1AS is a diagonal matrix.
Example: exercise 1(b), p. 340
Diagonalize the matrix A=5 6
22. In other words, find an
invertible matrix Sand a diagonal matrix Dsuch that
S1AS =Dor A=SDS1.
Solution. First find the eigenvalues and eigenvectors of A. The
vector 3
2is an eigenvector with eigenvalue 1, and the
vector 2
1is an eigenvector with eigenvalue 2. The matrix
S=3 2
21is the transition matrix from the eigenvector
basis to the standard basis, and the matrix S1AS is the
diagonal matrix 1 0
0 2.
pf2

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Math 304

Linear Algebra

Harold P. Boas

[email protected]

June 28, 2006

Highlights

From last time: I (^) application of eigenvectors to systems of differential equations Today: I (^) diagonalization of matrices and applications

When is a matrix diagonal?

Suppose a linear operator L on R^3 is represented in a basis

[ u 1 , u 2 , u 3 ] by the matrix

This means that L u 1 = 2 u 1 and L u 2 = 3 u 2 and L u 3 = 5 u 3. In other words, the basis vectors are eigenvectors of the operator L. A square matrix A is diagonalizable if the linear transformation x 7 → A x 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.

Example: exercise 1(b), p. 340

Diagonalize the matrix A =

. In other words, find an 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 vector

is an eigenvector with eigenvalue 1, and the

vector

is an eigenvector with eigenvalue 2. The matrix

S =

is the transition matrix from the eigenvector basis to the standard basis, and the matrix S−^1 AS is the diagonal matrix

Continuation

If A =

, find A^10.

Solution. Since S−^1 AS = D =

, and D^10 =

we have A^10 = SD^10 S−^1 =

− 3 + 212 − 6 + 3 × 211

2 − 211 4 − 3 × 210

More. Since the exponential function is given by a power series (ex^ = 1 + x + (^21)! x^2 + (^31)! x^3 + · · · + (^) n^1! xn^ + · · ·), we also have e (A := I + A + (^21)! A^2 + · · · = SeD^ S−^1 = 3 2 − 2 − 1

e 0 0 e^2

− 3 e + 4 e^2 − 6 e + 6 e^2 2 e − 2 e^2 4 e − 3 e^2

Application to differential equations

We have two ways to solve the system of differential equations y ′^ =

y.

From yesterday, we can write the general solution as y (t) = c 1 et

  • c 2 e^2 t

Alternatively, we can write the general solution as y (t) = etA

c 1 c 2

= SetD^ S−^1

c 1 c 2

− 3 et^ + 4 e^2 t^ − 6 et^ + 6 e^2 t 2 et^ − 2 e^2 t^4 et^ − 3 e^2 t

c 1 c 2

In the second form,

c 1 c 2

y 1 ( 0 ) y 2 ( 0 )