Find the symmetric matrix from eigen vectors

[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 ?

Thank you in advance .
Michael

 

[mfb]

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

 

[mihalisla]

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

 

[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 separated by ;

 

[mfb]

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.

 

[mihalisla]

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

 

[HallsofIvy]

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.

 

[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. . .

 

[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

#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

 

[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 ?