Question

1/4 y/x =12 and -1/y 2/x=6 form the given determinants, form the two simultaneous equations in x and y and solve them.

Answer 1

6.1 Introduction to Eigenvalues

Linear equations Ax D b come from steady state problems. Eigenvalues have their greatest

importance in dynamic problems. The solution of du=dt D Au is changing with time—

growing or decaying or oscillating. We can’t find it by elimination. This chapter enters a

new part of linear algebra, based on Ax D -

x. All matrices in this chapter are square.

A good model comes from the powers A; A2; A3;::: of a matrix. Suppose you need the

hundredth power A100. The starting matrix A becomes unrecognizable after a few steps,

and A100 is very close to Π:6 :6I :4 :4 :

-

:8 :3

:2 :7 -

:70 :45

:30 :55 -

:650 :525

:350 :475

-

-

-

-

:6000 :6000

:4000 :4000

A A2 A3 A100

A100 was found by using the eigenvalues of A, not by multiplying 100 matrices. Those

eigenvalues (here they are 1 and 1=2) are a new way to see into the heart of a matrix.

To explain eigenvalues, we first explain eigenvectors. Almost all vectors change di-

rection, when they are multiplied by A. Certain exceptional vectors x are in the same

direction as Ax. Those are the “eigenvectors”. Multiply an eigenvector by A, and the

vector Ax is a number -

times the original x.

The basic equation is Ax D -

x. The number -

is an eigenvalue of A.

The eigenvalue -

tells whether the special vector x is stretched or shrunk or reversed or left

unchanged—when it is multiplied by A. We may find -

D 2 or 1

2 or 1 or 1. The eigen-

value -

could be zero! Then Ax D 0x means that this eigenvector x is in the nullspace.

If A is the identity matrix, every vector has Ax D x. All vectors are eigenvectors of I .

All eigenvalues “lambda” are -

D 1. This is unusual to say the least. Most 2 by 2 matrices

have two eigenvector directions and two eigenvalues. We will show that det.A -

I / D 0.

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