# Diagonalize a matrix (help for exam)

1. Oct 22, 2008

### Gramsci

1. The problem statement, all variables and given/known data

Find a matrix that diagonalizes the following 2x2 matrix:

A= (1/2 , sqrt(3)/2
sqrt(3)/2,-1/2)

What will the diagonalizing matrix D be? What does D mean geometrically? What does A mean geometrically?

2. Relevant equations
-

3. The attempt at a solution

First I began with computing the characteristic equations determinant:
det A-lambda(call it x)= (1/2-x)(-1/2-x)-(sqrt3/2)(sqrt3/2)=X^2=-1
and since we haven't begun with complex eigenvalues yet:
x(x)=-1
Thus, x1=1 and x2=-1

Then I'm trying to compute the eigenvectors, but I seem to fail after I've added -1 I get:

3/2 , sqrt(3)/2
sqrt(3)/2, 1/2

after row reduction:
3, sqrt(3)
0, 0

Therefore, the eigenvector for -1 have to be (sqrt(3),3)
But according to my solutions manual, it is : 1, - sqrt 3

How do I count for the second one? All these square roots confuse me, any tips on how I handle them in row reduction?

/Gramsci

2. Oct 22, 2008

### HallsofIvy

You can't just declare the eigenvalues to be 1 and -1 because "haven't begun with complex eigenvalues yet"! What that tells you is that you have the wrong equation. (The eigenvalues of a symmetric matrix are always real.) (1/2- x)(-1/2- x)- (sqrt(3)/2)^2= x^2- 1/4- 3/4= x^2- 1= 0 which does have 1 and -1 as roots. Was that sheer luck?

You "added" -1? Since you were subtracting lambda, that means you subtracted 1. What you found is an eigenvector corresponding to 1, not -1. Also, any multiple of an eigenvector is also an eigenvector so there can be many different answer. If, for example, you divide what you got by sqrt(3), you get (1, sqrt(3)) as another eigenvector corresponding to the eigenvalue 1.

Go back to the originally matrix and subtract -1 rather than 1 (add 1 rather than -1). Row reduce that.

3. Oct 23, 2008

### Gramsci

Thanks for the help, now I get it. Thanks.