Math Is Fun Forum

deepz

Member

Registered: 2007-03-20

Posts: 10

Linear Algebra

Hi I need some help with the following question:

For what real values x is the symmetric real matrix A=

congruent to the matrix

and

I have put it into quadratic form and attempted to reduce it but am not getting anywhere. I get the 2x2 identity matrix except the (1,1) entry is x. As for the second congruent matrix it can be written as 2xy in quadratic form which can be written as (x+y)^2 - x^2 - y^2. That is as far as I could get. If anyone could help me by tomorrow morning I'd really appreciate it

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Linear Algebra

This might not be the method you're meant to use, but I went on wikipedia and it said that two symmetric matrices are congruent if they have the same amount of positive, negative and zero eigenvalues.

You find the eigenvalues of A by finding which values of λ make det(A-λI)=0.

In this case, det(A - λI) is given by λ² - (1+x+x³)λ +x.

Using the quadratic equation on this gives:

For the first part, the matrix is in diagonal form so the eigenvalues can be seen directly as 0 and 1. The only way for λ to equal 0 is for x to equal 0, so in this case there is only one solution.

For the second part, the eigenvalues turn out to be 1 and -1, so we need an x that gives λ_1 > 0 and λ_2 < 0.
Play with x and see what conditions on it make this happen.

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Why did the vector cross the road?
It wanted to be normal.