# Find the symmetric matrix from eigen vectors

1. Apr 24, 2013

### mihalisla

Hello to all of you,

Is there a way to get the matrix A=[a b c d] from the eigenvectors (orthogonal) matrix
H= sin(x) cos(x)
cos(x) -sin(x)
or to pose it differently to find a matrix that has these 2 eigenvectors ?

Michael

2. Apr 24, 2013

### Staff: Mentor

Just write the eigenvectors as columns of your matrix and watch what A*(1,0)^T does.

3. Apr 25, 2013

### mihalisla

^T is the transpose? and why [1,0]? thanks!

4. Apr 25, 2013

### mihalisla

I have made the assumption that matrix A should be symmetric because of the orthogonality of eigen vectors matrix! is this true? so matrix A = (a,b ;b, d) rows seperated by ;

5. Apr 25, 2013

### Staff: Mentor

Oh sorry, ignore my first post.
Do you have fixed eigenvalues for those eigenvectors? Otherwise, there is a lot of freedom: all multiplies of the identity matrix (including the identity matrix) have all vectors as eigenvectors, for example.
With fixed eigenvalues, I get 4 equations for 4 unknown parameters, so I would expect that there is a unique solution.

6. Apr 25, 2013

### mihalisla

no there are not fixed eigen values! how do I proceed?

7. Apr 25, 2013

### HallsofIvy

Staff Emeritus
If you don't know the eigenvalues, it is impossible to find the matrix. There are an infinite number of matrices having the same eigenvectors but different eigenvalues.
If D is the diagonal matrix, having the eigenvalues on the main diagonal and A is the matrix having the corresponding eigenvectors of the matrix as columns then $$ADA^{-1}$$ is the matrix having those eigenvalues and eigenvectors.

8. Apr 25, 2013

### mihalisla

I started from the eq A*H.1= lamda1*H.1 where H.1 is the first column of H (the A matrix of user HallsofIvy) .
Then I did the same for lamda2 (the second eigen value that is unknown) and I got lamda1=]a*sin(x)+b*cos(x)]/sin(x) and another value lamda1=[b*sin(x)+d*cos(x)]/cos(x).
Also two values for lamda2.
Should I try to get 4 equations for the four pairs of lamda1 and lamda2 doing A=H*L*H^-1????
my main objective is to set a,b,d from A as functios of sin and cos and not set numbers I suppose. . .

9. Apr 25, 2013

### mihalisla

Solved

Problem solved !!!
Matrix A=[a ,b]
[b, d]
From A*H.1=lamda1*H.1 I took
lamda1=(a*sin(x)+b*cos(x))/sin(x)
and lamda1=(b*sin(x)+d*cos)/cos(x)

From A*H.2=lamda2*H.2 I took
lamda2=(a*cos(x)-b*sin(x))/cos(x)
and lamda2=(d*sin(x)-b*cos(x))/sin(x)

For each set of lamda(i) i=1,2
i do next [A-lamda(i)]*H.i=0
All four sets of equations give me with free a ,d

b=((a-d)*cos(x)*sin(x))/(sin(x)^2-cos(x)^2)

It must be cos(x),sin(x) and sin(x)^2-cos(x)^2 not equal to zero

Thus matrix A takes the form A= a b=((a-d)*cos(x)*sin(x))/(sin(x)^2-cos(x)^2)

b=((a-d)*cos(x)*sin(x))/(sin(x)^2-cos(x)^2) d

(This is a square symmetric matrix )

In statistical program R i wrote a small script so as to validate the results

Code (Text):
#test1

x=(pi)
a=7
d=1
b=((a-d)*cos(x)*sin(x))/(sin(x)^2-cos(x)^2)

l1=(a*sin(x)+b*cos(x))/sin(x)
l2=(a*cos(x)-b*sin(x))/cos(x)

A=matrix(c(a,b,b,d),2,2)
H=matrix(c(sin(x),cos(x),cos(x),-sin(x)),2,2)
L=matrix(c(l1,0,0,l2),2,2)
#it must be A*H=H*L where L is the diagonial matrix with lamda1 and lamda2 as l1 and l2 #respectively

A%*%H
H%*%L

#here I get A from H*L*H^-1
H%*%L%*%ginv(H)
#here I get A from H*L*H^T
H%*%L%*%t(H)
A

Last edited: Apr 25, 2013
10. Apr 25, 2013

### mihalisla

I have one more but i ll post it in new thread !
P.S I m new here so posting the solution is good , isn't it ???