# Help on a matrix problem.

• Oct 28th 2010, 04:48 AM
taurine330
Help on a matrix problem.
'The Matrix

A = ( -1 1 -1)
( 0 c 0)
(3+2c-c^2 b 3)

can only be diagonalised for some values of the parameters b and c . Construct a table
showing the values of b and c for which diagonalisation is possible. You must show

For this I have done the following:

eigenvectors -1

P(ƛ) = -(c -ƛ)((-1-ƛ)(3-ƛ)+3+2c-c^2)

For what I finally got

c-ƛ =0 -> ƛ =c

Then ƛ^2 -2ƛ +2C -ƛ^2=0

and then getting the value for ƛ

ƛ = 1+-(1-c)

-----

Is this right?
And what should I do next with this problem?

Thank you for your time.
• Oct 28th 2010, 04:52 AM
Ackbeet
Your work so far is correct. At least, those are the eigenvalues I get. You'll notice there are only two eigenvalues. What are the algebraic multiplicities of those two eigenvalues?
• Oct 28th 2010, 05:03 AM
taurine330
Quote:

Originally Posted by Ackbeet
Your work so far is correct. At least, those are the eigenvalues I get. You'll notice there are only two eigenvalues. What are the algebraic multiplicities of those two eigenvalues?

So I need to find the value 'k' ?
• Oct 28th 2010, 05:06 AM
Ackbeet
If by 'k' you mean the integer $k$ such that

$\det(A-\lambda I)=(\lambda-2+c)(\lambda-c)^{k}F,$

I would agree with you. You haven't defined k otherwise, so I don't know what it is yet.
• Oct 28th 2010, 05:19 AM
taurine330
Quote:

Originally Posted by Ackbeet
If by 'k' you mean the integer $k$ such that

$\det(A-\lambda I)=(\lambda-2+c)(\lambda-c)^{k}F,$

I would agree with you. You haven't defined k otherwise, so I don't know what it is yet.

Yea, that's what I meant. I kinda understand this. But not sure if I know what should I do next.
• Oct 28th 2010, 05:25 AM
Ackbeet
Well, for the determinant that defines the characteristic equation, I get

$-(c-\lambda)^{2}(\lambda-2+c).$

So what is $k$?
• Oct 28th 2010, 09:10 AM
taurine330
Thank you.

When the integer k is acquired is there something that still that needs to be done for the problem?
• Oct 28th 2010, 09:12 AM
Ackbeet
Yes. Is there a criterion you know of whereby you can say that a matrix is diagonalizable or not?
• Oct 28th 2010, 09:13 AM
taurine330
Quote:

Originally Posted by Ackbeet
Yes. Is there a criterion you know of whereby you can say that a matrix is diagonalizable or not?

It says:

"The Matrix can only be diagonalised for some values of the parameters b and c. Construct a table showing the values of b and c for which diagonalisation is possible."
• Oct 28th 2010, 09:15 AM
Ackbeet
That's in your problem statement. But you haven't answered my question.

Independent of this problem, what criterion do you know of that will tell you whether a matrix is diagonalizable or not?
• Oct 28th 2010, 09:20 AM
taurine330
Quote:

Originally Posted by Ackbeet
That's in your problem statement. But you haven't answered my question.

Independent of this problem, what criterion do you know of that will tell you whether a matrix is diagonalizable or not?

Is this when there exists an invertible matrix P such that P −1AP ?
The matrix can be written in form A=PDP^(-1)
• Oct 28th 2010, 09:58 AM
Ackbeet
That is the definition of diagonalizable, I grant you. However, it's not much use in the present circumstance, I'm afraid. Have you seen this one:

A matrix A is diagonalizable if and only if, for every eigenvalue of A, its geometric multiplicity equals its algebraic multiplicity.

You always know that the geometric multiplicity of an eigenvalue is less than or equal to the algebraic multiplicity. Here, we're saying that they have to be equal for all the eigenvalues. If that happens, you can diagonalize the matrix.

So, how can you use this idea to find out when your matrix is diagonalizable?
• Oct 28th 2010, 10:49 AM
taurine330
Quote:

Originally Posted by Ackbeet
That is the definition of diagonalizable, I grant you. However, it's not much use in the present circumstance, I'm afraid. Have you seen this one:

A matrix A is diagonalizable if and only if, for every eigenvalue of A, its geometric multiplicity equals its algebraic multiplicity.

You always know that the geometric multiplicity of an eigenvalue is less than or equal to the algebraic multiplicity. Here, we're saying that they have to be equal for all the eigenvalues. If that happens, you can diagonalize the matrix.

So, how can you use this idea to find out when your matrix is diagonalizable?

Is that then eigenvalue c has to equal the value k?

c= k

Cheers
• Oct 28th 2010, 03:24 PM
Ackbeet
No, no. Try to solve the equation $(A-cI)x=0,$ and see what you get. You're trying to find eigenvectors now, and you want to know when you need two vectors, and when you need one.