I'm afraid it is eigen space related esamani-jan.
Thanks for the offer Flint-jan.
Here is the problem.
Let A be a 3x3 matrix. The eigenvalues of A are L such that:
Ax = Lx (Eq. 1)
where x are the eigenvectors.
The way we find the eigenvalues is by solving the characteristic polynomial of A defined as:
det(A- LI)=0 (Eq. 2)
where "det" denotes the matrix determinant, L are eigenvalues of A and I is the identify matrix.
The characteristic polynomial for 3x3 matrices turns out to be the third degree polynomial given by:
L^3 - tr(A)*L^2 - 0.5*[tr(A^2) - tr(A)*tr(A)]*L - det(A) = 0 (Eq. 3)
where "tr" denotes the trace of the matrix.
Having said all this, let me now present the generalized eigenvalue problem. Here we are looking for the generalized eigenvalues of A that would satisfy the following:
Ax = LBx (Eq. 4)
The difference between this and Eq. 1 is the presence of the (3x3) matrix B on the right hand side of Eq. 4. So the generalized eigenvalue problem is, given matrices A and B, find the eigenvalues L and eigenvectors x that satisfy Eq. 4.
Again to solve for L we have to find the roots of the following characteristic polynomial:
del( A - LB) = 0 (Eq. 5)
Note that this is similar toe Eq. 2, with the exception that here matrix B replaces the identify matrix I in Eq. 2.
So FINALLY, here is my question, how can I express Eq. 5 in the form of Eq. 3, that is, how can I explicitly write Eq. 5 as a third degree polynomial of L. On other words, I would like to express Eq. 5 as something like:
a3*L^3 + a2*L^2 + a1*L + a0 = 0 (Eq. 6)
and be able to express the coefficients of this polynomial (a0, a1, a2, a3) explicitly in terms of matrix A and B.
In the regular eigenvalue problem, I was able to write the characteristic polynomial (Eq. 2) as in (Eq. 6) by setting:
a3 = 1
a2 = -tr(A)
a1 = -0.5*[tr(A^2)-tr(A)*tr(A)]
and
a0 = -det(A)
I would like to do the same thing for the GENERALIZED eigenvalue problem. That is, to obtain the coefficients of the characteristic polynomial of Eq. (5) explicitly in terms of matrices A and B.
Thanks for the offer Flint-jan.
Here is the problem.
Let A be a 3x3 matrix. The eigenvalues of A are L such that:
Ax = Lx (Eq. 1)
where x are the eigenvectors.
The way we find the eigenvalues is by solving the characteristic polynomial of A defined as:
det(A- LI)=0 (Eq. 2)
where "det" denotes the matrix determinant, L are eigenvalues of A and I is the identify matrix.
The characteristic polynomial for 3x3 matrices turns out to be the third degree polynomial given by:
L^3 - tr(A)*L^2 - 0.5*[tr(A^2) - tr(A)*tr(A)]*L - det(A) = 0 (Eq. 3)
where "tr" denotes the trace of the matrix.
Having said all this, let me now present the generalized eigenvalue problem. Here we are looking for the generalized eigenvalues of A that would satisfy the following:
Ax = LBx (Eq. 4)
The difference between this and Eq. 1 is the presence of the (3x3) matrix B on the right hand side of Eq. 4. So the generalized eigenvalue problem is, given matrices A and B, find the eigenvalues L and eigenvectors x that satisfy Eq. 4.
Again to solve for L we have to find the roots of the following characteristic polynomial:
del( A - LB) = 0 (Eq. 5)
Note that this is similar toe Eq. 2, with the exception that here matrix B replaces the identify matrix I in Eq. 2.
So FINALLY, here is my question, how can I express Eq. 5 in the form of Eq. 3, that is, how can I explicitly write Eq. 5 as a third degree polynomial of L. On other words, I would like to express Eq. 5 as something like:
a3*L^3 + a2*L^2 + a1*L + a0 = 0 (Eq. 6)
and be able to express the coefficients of this polynomial (a0, a1, a2, a3) explicitly in terms of matrix A and B.
In the regular eigenvalue problem, I was able to write the characteristic polynomial (Eq. 2) as in (Eq. 6) by setting:
a3 = 1
a2 = -tr(A)
a1 = -0.5*[tr(A^2)-tr(A)*tr(A)]
and
a0 = -det(A)
I would like to do the same thing for the GENERALIZED eigenvalue problem. That is, to obtain the coefficients of the characteristic polynomial of Eq. (5) explicitly in terms of matrices A and B.
I think I make the question somewhat convoluted. Let me put it more simply.
Let:
A be a 3x3 matrix
L be a scalar
I be 3x3 identify matrix
The characteristic polynomial of A:
det(A - LI) = 0
is a third degree polynomial:
a3*L^3 + a2*L^2 + a1*L + a0 = 0
where
a3 = 1
a2 = -tr(A)
a1 = -0.5*[tr(A^2) - tr(A)*tr(A)]
a0 = det(A).
Now my question. Let B be another 3x3 matrix.
I would like to express
det(A - LB) = 0
as
a3*L^3 + a2*L^2 + a1*L + a0 = 0
In this case, the question is what are a3, a2, a1, a0 in terms of matrices A and B?
Let:
A be a 3x3 matrix
L be a scalar
I be 3x3 identify matrix
The characteristic polynomial of A:
det(A - LI) = 0
is a third degree polynomial:
a3*L^3 + a2*L^2 + a1*L + a0 = 0
where
a3 = 1
a2 = -tr(A)
a1 = -0.5*[tr(A^2) - tr(A)*tr(A)]
a0 = det(A).
Now my question. Let B be another 3x3 matrix.
I would like to express
det(A - LB) = 0
as
a3*L^3 + a2*L^2 + a1*L + a0 = 0
In this case, the question is what are a3, a2, a1, a0 in terms of matrices A and B?
Now onto my another problem, given L is a generalized eigenvalue of Ax=LBx, I would like to find the derivative of L with respect to any given element of b (say bij). What I have in mind is to write the characteristic polynomial:
L^3 + a2*L^2 + a1*L + a0 = 0
Differential both sides wrt to bij to get:
3L^2*dL/dbij + a2*2L*dL/dbij + da2/dbij * L^2 + a1*dL/dbij + da1/dbij * L + da0/dbij = 0
Then solve for dL/dbij.
But given the expressions for a0, a1, a2 involving trace and determinants of B^{-1}A I think things are going to get pretty hairy.
L^3 + a2*L^2 + a1*L + a0 = 0
Differential both sides wrt to bij to get:
3L^2*dL/dbij + a2*2L*dL/dbij + da2/dbij * L^2 + a1*dL/dbij + da1/dbij * L + da0/dbij = 0
Then solve for dL/dbij.
But given the expressions for a0, a1, a2 involving trace and determinants of B^{-1}A I think things are going to get pretty hairy.
You don't know the answer mizani kanal 2? Reminds me of my uncle at the beginning of revolution when "isms" were used like noghl o nabat. We were at his house and TV was on and reporter would say blah blah blah "Imperialism". One of the guests asked my uncle "What does Imperialism mean?" and he answered. then reporter said blah blah blah "communism", the question came "What does it mean?" and my uncle answered. goes on and on until reporter said "existentialism". I now asked my uncle what it meant? He didn't know the answer but didn't want to say it, so he turned to me and said"be quiet I am watching the news (or Heees Akhbar!!!")
Generally we have (beware, my notation is not mathematically strict!): Ax = Lx ---> det(A-IL)=0 gives ch. pol.
Now we have (assuming B is invertible, i.e. B^-1 exists!): Ax = LBx ---> B^(-1)Ax = Lx, so yeah, det(B^(-1) - IL)=0 gives eigenvalues.
det(A-BL)=0 is also correct, but this is a general expression. If B is invertible, it reduces to the above... if you text says B is invertible, then go with the above (which is much easier!), otherwise you have to do some crunching.
Now we have (assuming B is invertible, i.e. B^-1 exists!): Ax = LBx ---> B^(-1)Ax = Lx, so yeah, det(B^(-1) - IL)=0 gives eigenvalues.
det(A-BL)=0 is also correct, but this is a general expression. If B is invertible, it reduces to the above... if you text says B is invertible, then go with the above (which is much easier!), otherwise you have to do some crunching.
In ch. pol.
L^3 - tr(A)*L^2 - 0.5*[tr(A^2) - tr(A)*tr(A)]*L - det(A) = 0
when I substitute B^{-1}A for A, I obtain:
L^3 - tr(B^{-1}A)*L^2 - 0.5*{tr[(B^{-1}A)^2] - tr(B^{-1}A)*tr(B^{-1}A)}*L - det(B^{-1}A) = 0
Then, interestingly, when I multiply both sides by det(B) I obtain:
det(B)*L^3 - tr[ adj(B)*A]*L^2 - 0.5*det(B)*{tr[(B^{-1}A)^2] - tr(B^{-1}A)*tr(B^{-1}A)}*L - det(A) = 0
where adj(B) denotes the adjoint of B and I used the fact that adj(B) = B^{-1} * det(B).
Furthermore, multiplying the coefficient of L byt det(B)/det(B), we obtain:
det(B)*L^3 - tr[ adj(B)*A]*L^2 - 0.5*(1/det(B))*{tr[(adj(B)*A)^2] - tr(adj(B)*A)*tr(adj(B)*A)}*L - det(A) = 0
This is the form I was looking for fromt he beginning. The beauty of it is that by multiplying by det(B), I got rid of the terms involving B^{-1} and replaced it by adj(B). This makes differentiation much easier.
Furthermore, in the problem that I am dealing with, I failed to tell you that both A and B are symmetric matrices. If you used this fact, believe or not, the whole ch. pol. just reduces to the following expression:
det(B)*L^3 - tr[adj(B)*A]*L^2 + tr[B*adj(A)] *L - det(A) = 0
That awful looking coefficient of the L term just reduces to tr[B*adj(A)] !!!!!
I find this a beautiful result. Just look at the symmetry of that equation.
It also makes my differentiation problem much easier. Notice that:
derivative of det(A) wrt A equals adj(A) and
derivative of tr[ adj(B)*A ] wrt A equals adj(B).
I am still working to determined the expression for the derivative of tr[ B*adj(A) ] wrt A.
So I am very close to finding the derivatives of L wrt A which is my final goal.
L^3 - tr(A)*L^2 - 0.5*[tr(A^2) - tr(A)*tr(A)]*L - det(A) = 0
when I substitute B^{-1}A for A, I obtain:
L^3 - tr(B^{-1}A)*L^2 - 0.5*{tr[(B^{-1}A)^2] - tr(B^{-1}A)*tr(B^{-1}A)}*L - det(B^{-1}A) = 0
Then, interestingly, when I multiply both sides by det(B) I obtain:
det(B)*L^3 - tr[ adj(B)*A]*L^2 - 0.5*det(B)*{tr[(B^{-1}A)^2] - tr(B^{-1}A)*tr(B^{-1}A)}*L - det(A) = 0
where adj(B) denotes the adjoint of B and I used the fact that adj(B) = B^{-1} * det(B).
Furthermore, multiplying the coefficient of L byt det(B)/det(B), we obtain:
det(B)*L^3 - tr[ adj(B)*A]*L^2 - 0.5*(1/det(B))*{tr[(adj(B)*A)^2] - tr(adj(B)*A)*tr(adj(B)*A)}*L - det(A) = 0
This is the form I was looking for fromt he beginning. The beauty of it is that by multiplying by det(B), I got rid of the terms involving B^{-1} and replaced it by adj(B). This makes differentiation much easier.
Furthermore, in the problem that I am dealing with, I failed to tell you that both A and B are symmetric matrices. If you used this fact, believe or not, the whole ch. pol. just reduces to the following expression:
det(B)*L^3 - tr[adj(B)*A]*L^2 + tr[B*adj(A)] *L - det(A) = 0
That awful looking coefficient of the L term just reduces to tr[B*adj(A)] !!!!!
I find this a beautiful result. Just look at the symmetry of that equation.
It also makes my differentiation problem much easier. Notice that:
derivative of det(A) wrt A equals adj(A) and
derivative of tr[ adj(B)*A ] wrt A equals adj(B).
I am still working to determined the expression for the derivative of tr[ B*adj(A) ] wrt A.
So I am very close to finding the derivatives of L wrt A which is my final goal.