Thread Subject: Eigenvalues Problem?

Hello... Good Day.
I Have A Problem With Eigenvalues polinomial characteristic.

I've read some books to get eigenvalues.

If A=2x2:
a b
c d

then

polinomial is: lambda^2 - (a+b)lambda + det(A)

So

If A=3x3

a b c
d e f
g h i

polinomial is: -lambda^3 + (a+b+c) lambda^2 -
(db+gc+hf-ae-ei) lambda + det(A)

so polinomial of eigenvalues is:
(-1)^n lambda^n + (-1)^(n-1) S1 lambda^(n-1) + (-1)^(n-2) Sk
lambda^(k) + Sn

Where:
n=matrix dimension
s1=trace(A)
Sn=det(A)

I wanna ask how to get Sk?

If I have 4 dimension:

a b c d
e f g h
i j k l
m n o p

Thanks.

On 30 Dec, 08:31, "Michael Nikki" <ni...@yahoo.com> wrote:
> Hello... Good Day.
> I Have A Problem With Eigenvalues polinomial characteristic.
>
> I've read some books to get eigenvalues.
>
> If A=2x2:
> a b
> c d
>
> then
>
> polinomial is: lambda^2 - (a+b)lambda + det(A)
>
> So
>
> If A=3x3
>
> a b c
> d e f
> g h i
>
> polinomial is: -lambda^3 + (a+b+c) lambda^2 -
> (db+gc+hf-ae-ei) lambda + det(A)
>
> so polinomial of eigenvalues is:
> (-1)^n lambda^n + (-1)^(n-1) S1 lambda^(n-1) + (-1)^(n-2) Sk
> lambda^(k) + Sn
>
> Where:
> n=matrix dimension
> s1=trace(A)
> Sn=det(A)
>
> I wanna ask how to get Sk?
>
> If I have 4 dimension:
>
> a b c d
> e f g h
> i j k l
> m n o p
>
> Thanks.

Michel,

Two points:

1) you are getting your traces wrong - they
should be the sum of the diagonal elements

so If A=2x2:
> a b
> c d

then trace A =a+d not a+b, as you write.

2) generally, the characteristic polynomial is
obtained from det(A-lambda*I)
where I is the identity matrix, e.g. in the 2x2 case
I is

1 0
0 1

So you need to read up how to work out
determinants.

Hope this helps.

Inf.

Inf <infinitysquared@gmail.com> wrote in message
> Michel,
>
> Two points:
>
> 1) you are getting your traces wrong - they
> should be the sum of the diagonal elements
>
> so If A=2x2:
> > a b
> > c d
>
> then trace A =a+d not a+b, as you write.

Yeah, I Apologize from that.

>
> 2) generally, the characteristic polynomial is
> obtained from det(A-lambda*I)
> where I is the identity matrix, e.g. in the 2x2 case
> I is
>
> 1 0
> 0 1
>
> So you need to read up how to work out
> determinants.
>
> Hope this helps.
>
> Inf.

You mean that when getting lambda 1-n, I can get from
det(A-lambda*I)?

So how to get Sk from

(-1)^n lambda^n + (-1)^(n-1) S1 lambda^(n-1) + (-1)^(n-2) Sk
> > lambda^(k) + Sn

It says sub matrix from A (k x k). What the mean of sub
matrix? Sorry I'm very newbie from this. Thanks for your answer.

"Michael Nikki" <niqmk@yahoo.com> wrote in message <fl7qqb$1s7
$1@fred.mathworks.com>...
> ........
> So how to get Sk from
>
> (-1)^n lambda^n + (-1)^(n-1) S1 lambda^(n-1) + (-1)^(n-2) Sk
> > > lambda^(k) + Sn
>
> It says sub matrix from A (k x k). What the mean of sub
> matrix? Sorry I'm very newbie from this. Thanks for your answer.
--------
  By expanding the expression, det(A-lambda*I), and collecting equal powers
of lambda, it can be seen that each of your coefficients, Sk, is the "sum of all
the principal minor determinants of A with k rows", to quote one website.
That is, Sk is the sum of all the determinants with k rows and k columns that
can be obtained by eliminating n-k of the diagonal elements of A, along with
the corresponding row and column of each. For example, with your 3 x 3
determinant

 A =
 a b c
 d e f
 g h i

the value of S2 would be the sum of the three determinants

 det([a b;d e]) + det([a c;g i]) + det([e f;h i]) =
 (a*e-b*d)+(a*i-c*g)+(e*i-f*h) ,

which agrees with your earlier expression (if you correct the missing ai.)
These determinants are all the possible minor determinants of A with one
(3-2) diagonal element and the associated row and column removed.

  As you can see, for a general k there will be nCk (the number of ways of
choosing k out of n things) different possible minor determinants forming the
sum in Sk. I am not aware of any compact formula for such a sum.
Computing the characteristic polynomial for a general square matrix of large
order is therefore a non-trivial task.

Roger Stafford