I have an equation of the form A*X = X*A', where A and X are real, square matrices (3x3 in that case) and A is known and A' represent the transpose of A. How to solve for X using MATLAB? (up to a scale factor)
This is a Sylvester equation. However, it is singular because the eigenvalues of A and A' are the same. But you can use the formulae
[I⊗A-A'⊗I]X(:)=C(:):
m=kron(eye(3),a)+kron(-a,eye(3))
v=null(m)
x1=reshape(v(:,1),[3 3])
x2=reshape(v(:,2),[3 3])
x3=reshape(v(:,3),[3 3])
Now the solution is span{x1,x2,x2}, i.e. any matrix of the form
b x1 + c x2 +d x3, where b,c,d are any real numbers
I don't think Matlab has facilities for symbolic algebra.
If you expand A and X, and work through the expression, you obtain an 3x3 matrix with equation in several unknowns, all of which are zero. You then solve that.
But I don't think Matlab allows you to set a matrix to a symbol, rather than an value and expand it for you. For this simple case, you could easily write such a function, that multiples a string matrix by a numerical matrix. The snag is it's hard to scale it up to the general case without throwing the entire Maple / Mathematica engine at it.
Related
In Matlab, I'm trying to solve for the energies and eigenstates of a Hamiltonian matrix which has a highly degenerate set of eigenvectors. The matrix is a 55x55 hermitian matrix, and when I call either eig or schur to do the diagonalization I find that some (but not all) of the eigenvectors are the "wrong" linear combinations within each degenerate subspace. What I mean by "wrong" is that there are additional constraints in the problem. In this case, there is a good quantum number, M, which I want to preserve by not allowing states with different M values to be mixed--- but that mixing is exactly what I see when I run the code. Is there a way to tell Matlab to diagonalize the matrix while simultaneously maintaining the eigenvectors of another operator?
you can use diag to diagonalize a matrix and [eig_vect,eig_val] = eig(A) to give you eigenvectors.
I don't know matlab well enough to know whether there is a routine for this this, but here's how to do it algorithmically:
First diagonalise H, as you do now. Then for each degenerate eigen-space V, diagonalise the restriction of C to V, and use this diagonalisation to compute simulaneous diagonalisations of C and H
In more detail:
I assume you have an operator C that commutes with your Hamiltonian H. If V is the eigen-space of H for a particular (degenerate) eigen value, and you have a basis x[1] .. x[n] of V , then for each i, Cx[i] must be in V, and so we can expand Cx[i] in terms of the x[], and get a matrix representation C^ of the restriction of C to V, that is we compute
C^[k,j] = <x[k]|C*x[j]> k,j =1 .. n
Diagonalise the matrix C^, getting
C^ = U*D*U*
Then for each row (r1,..rn) of U* we can form
chi = Sum{ j | r[j]*x[j]}
A little algebra shows that this is an eigenvector of C, and also of H
I have a matrix that is a function of some parameter A=A(x). I would like to find the points x where this matrix becomes singular. Example (I have a large matrix though):
syms x
A=[x sin(x); cos(x^2) 2.5];
So far I have been symbolically computing the determinant of the matrix and then used fzero or newtzero to find the roots of that characteristic equation. I.e.
detA = det(A);
fzero(matlabFunction(detA),startingGuess)
Then I found this: How to find out if a matrix is singular?, where it is advocated to not use the determinant under any circumstances.
Indeed the symbolic determinant calculation is terribly slow. However I tried to use rank(A) instead as suggested in the link and it does not seem to work for symbolic matrices.
Is there any way to implement the suggestions in the link for finding the roots of a characteristic equation of a matrix that is given symbolically?
A possible approach would be the following: a square matrix A is singular if and only if the homogeneous linear (with respect to the vector y) system A*y = 0 has nontrivial solutions y <> 0 (which is equivalent to det(A) = 0 and rank(A) = 0 among others. So a more or less standard, as I recall from the past, technique to compute such points x is to solve the nonlinear system
A(x)*y = 0 (1)
||y|| = 1 (2)
This way you can compute a point x* and a vector y* such that A(x*) is singular and y* is an eigenvector corresponding to the zero eigenvalue of A(x*).
If I remember correctly, you can also solve the somewhat easier system
A(x)*y = 0 (1)
<y,c> = 1 (2a)
where c is "almost" any nonzero random vector (normalize it to 1 to avoid numerical problems).
As a matter of fact there is an enormous bibliography on the subject - you can look for saddle-node bifurcation computations (in case A(x) is the Jacobian of a vector field), or for "distance to instability".
From a discussion with Ander Biguri it seems that the determinant is actually a perfectly fine method of approaching this problem. The problem seems to be to solve the final equation in a stable manner, which would be a different question.
I have the following equation:
I want to do a exponential curve fitting using MATLAB for the above equation, where y = f(u,a). y is my output while (u,a) are my inputs. I want to find the coefficients A,B for a set of provided data.
I know how to do this for simple polynomials by defining states. As an example, if states= (ones(size(u)), u u.^2), this will give me L+Mu+Nu^2, with L, M and N being regression coefficients.
However, this is not the case for the above equation. How could I do this in MATLAB?
Building on what #eigenchris said, simply take the natural logarithm (log in MATLAB) of both sides of the equation. If we do this, we would in fact be linearizing the equation in log space. In other words, given your original equation:
We get:
However, this isn't exactly polynomial regression. This is more of a least squares fitting of your points. Specifically, what you would do is given a set of y and set pair of (u,a) points, you would build a system of equations and solve for this system via least squares. In other words, given the set y = (y_0, y_1, y_2,...y_N), and (u,a) = ((u_0, a_0), (u_1, a_1), ..., (u_N, a_N)), where N is the number of points that you have, you would build your system of equations like so:
This can be written in matrix form:
To solve for A and B, you simply need to find the least-squares solution. You can see that it's in the form of:
Y = AX
To solve for X, we use what is called the pseudoinverse. As such:
X = A^{*} * Y
A^{*} is the pseudoinverse. This can eloquently be done in MATLAB using the \ or mldivide operator. All you have to do is build a vector of y values with the log taken, as well as building the matrix of u and a values. Therefore, if your points (u,a) are stored in U and A respectively, as well as the values of y stored in Y, you would simply do this:
x = [u.^2 a.^3] \ log(y);
x(1) will contain the coefficient for A, while x(2) will contain the coefficient for B. As A. Donda has noted in his answer (which I embarrassingly forgot about), the values of A and B are obtained assuming that the errors with respect to the exact curve you are trying to fit to are normally (Gaussian) distributed with a constant variance. The errors also need to be additive. If this is not the case, then your parameters achieved may not represent the best fit possible.
See this Wikipedia page for more details on what assumptions least-squares fitting takes:
http://en.wikipedia.org/wiki/Least_squares#Least_squares.2C_regression_analysis_and_statistics
One approach is to use a linear regression of log(y) with respect to u² and a³:
Assuming that u, a, and y are column vectors of the same length:
AB = [u .^ 2, a .^ 3] \ log(y)
After this, AB(1) is the fit value for A and AB(2) is the fit value for B. The computation uses Matlab's mldivide operator; an alternative would be to use the pseudo-inverse.
The fit values found this way are Maximum Likelihood estimates of the parameters under the assumption that deviations from the exact equation are constant-variance normally distributed errors additive to A u² + B a³. If the actual source of deviations differs from this, these estimates may not be optimal.
is it possible to solve below equation in matlab?
A*X+B*exp(X)=C
A, B are square and constant matrices. C is a constant and column matrix.
X is a column matrix which should be found.( exp() acts element by element on X).
If you are looking for a numeric method, you might want to try fsolve
X = fsolve( #(x) A*x + B*exp(x) - C, x0 );
Due to the non-linear nature of the problem you need to provide an initial guess x0 - the quality of which can affect the performance of the solver.
I have to solve the following equation in matlab but I have some requirement step which I must follow.
x+y=17
2x-y=10
"write the system of equation on Matrix form,
AX=B, and plug in the matrix A and the vector B in Matlab. Solve the system
of equations by calculating the inverse A-1, and then the product A-1 B using
Matlab. Also, report the determinant of A in each question".
Writing this as a matrix equation:
A * X = B
You have the following matrices:
A = [1 1; 2 -1];
B = [17; 10];
And you are looking for X. For this you need the matlab \ operator. I am going to leave it as an exercise for the student to figure out how to use that operator with this matrix and vector combination in order to produce X which is the solution of the equation.