Problem about RootOf in Singular Value Decomposition

Hello,

I have a problem about getting rid of the RootOf function that appears as a result of a singular value decomposition (SVD). I'd appreciate your help.

I obtain a vector of singular values of a 4x4 matrix L by "sigma=svd(L)". L has only one variable, which is "kx". The vector of singular values is given below. What Matlab does is to add an additional variable "z1" to the result. As a result, I have two variables, kx and z1. My question is how I can obtain a vector of singular values without a "z1" variable?

In addition, there is a "[1]^(1/2)" term at the end of each row. What does that point out?

The result of sigma=svd(L) is:

sigma=
RootOf(256*kx^2*z1^6*conj((- kx^2 + 9/4)^(1/2))^2)[1]^(1/2)
RootOf(256*kx^2*z1^6*conj((- kx^2 + 9/8)^(1/2))^2)[1]^(1/2)
RootOf(256*kx^2*z1^6*conj((- kx^2 + 9/16)^(1/2))^2)[1]^(1/2)
RootOf(256*kx^2*z1^6*conj((- kx^2 + 9/24)^(1/2))^2)[1]^(1/2)

Thanks in advance,

Erdem

Answer 1

[Star Strider]

The z1 is a dummy variable the Symbolic Math Toolbox inserts in some situations. Without seeing your code, it is difficult to say what it could mean. The RootOf may have to do with the symbolic engine assuming that kx could be complex. Declaring it real in your syms or sym statement could clear that up. No guarantees.

[Erdem]

Hello,

Thank you for the answer. Attached I am sending the code. It is a quite short code. L in this code is a 6x6 matrix, all the other procedure is the same. I'd appreciate your help.

clear all
syms kz1 kz2 kz3 kx

d2=50e-9; eps1=2.25; eps2=2.25; eps3=2.25; 
w=10; c=30;

kz1 = sqrt((((w./c).^2).*eps1)-kx.^2);
kz2 = sqrt((((w./c).^2).*eps2)-kx.^2);
kz3 = sqrt((((w./c).^2).*eps3)-kx.^2);

L=[0 1 0 0 0 0;...
   1 1 -1 -1 0 0;... 
   -eps1/kz1 eps1/kz1 eps2/kz2 -eps2/kz2 0 0;
   0 0 exp(kz2*d2) exp(-kz2*d2) -exp(kz3*d2) -exp(-kz3*d2);...
   0 0 (-eps2/kz2)*exp(kz2*d2) (eps2/kz2)*exp(-kz2*d2) (eps3/kz3)*exp(kz3*d2) (-eps3/kz3)*exp(-kz3*d2);...
   0 0 0 0 1 0];

sigma = svd(L)

Erdem

[Star Strider]

Each of the 6 results in sigma are 6-th degree polynomials in the dummy variable z. This is an intractable problem for the symbolic solver.

I am not certain what you want to do, but it is best that you write a function with kx as the only input argument, and then evaluate it numerically.

[Erdem]

Thanks for the answer.

Actually my aim is to multiply all of the entries of sigma with each other (i.e. sigma(1)*sigma(2)*...*sigma(6)). Since the resulting multiplication will have no explicit solution, I want to use "fminsearch" to solve for kx. Therefore kx has to stay in the expression as it is. I cannot assign it a specific value. It is the potential solution of the expression.

[Star Strider]

My pleasure!

Using fminsearch should be relatively easy. In fact, the Symbolic Math Toolbox function matlabFunction will convert your matrix to an anonmymous function for you:

Erdem = @(kx) reshape([0.0,1.0,1.0./sqrt(-kx.^2+1.0./4.0).*(-9.0./4.0),0.0,0.0,0.0,1.0,1.0,1.0./sqrt(-kx.^2+1.0./4.0).*(9.0./4.0),0.0,0.0,0.0,0.0,-1.0,1.0./sqrt(-kx.^2+1.0./4.0).*(9.0./4.0),exp(sqrt(-kx.^2+1.0./4.0).*5.0e-8),exp(sqrt(-kx.^2+1.0./4.0).*5.0e-8).*1.0./sqrt(-kx.^2+1.0./4.0).*(-9.0./4.0),0.0,0.0,-1.0,1.0./sqrt(-kx.^2+1.0./4.0).*(-9.0./4.0),exp(sqrt(-kx.^2+1.0./4.0).*(-5.0e-8)),exp(sqrt(-kx.^2+1.0./4.0).*(-5.0e-8)).*1.0./sqrt(-kx.^2+1.0./4.0).*(9.0./4.0),0.0,0.0,0.0,0.0,-exp(sqrt(-kx.^2+1.0./4.0).*5.0e-8),exp(sqrt(-kx.^2+1.0./4.0).*5.0e-8).*1.0./sqrt(-kx.^2+1.0./4.0).*(9.0./4.0),1.0,0.0,0.0,0.0,-exp(sqrt(-kx.^2+1.0./4.0).*(-5.0e-8)),exp(sqrt(-kx.^2+1.0./4.0).*(-5.0e-8)).*1.0./sqrt(-kx.^2+1.0./4.0).*(-9.0./4.0),0.0],[6,6]);

Q = Erdem(1)        % Test at kx = 1

I named it Erdem, name it whatever you want. (I used the Q assignment as a test to be sure it returns a matrix.) All you need to do is to create an anonymous function using it as an argument to svd and you can then use it with fminsearch. The optimisation functions want a single scalar output from the argument function, so since svd returns a vector of singular values, I’ll leave it to you to decide how you want to write that function. If you are matching it to a known vector and you want to solve for a value of kx that creates that vector, use the norm function on the difference. (I thank Matt J for that valuable insight.)

[Alfonso Nieto-Castanon]

In case this helps simplify things, I believe the product of sigma (product of singular values):

 prod(svd(L))

can also be expressed more simply as:

 sqrt(det(L'*L))

This should allow you to express analytically your measure of interest.

[Erdem]

This is a clever idea! Thanks Star Strider. Thank to you too Alfonso.

Star Strider, how can I create an anonymous function as an argument to svd? If it is simple to show, is it possible for you to write it down? By the way, does svd accept to be called by an anonymous function?

Actually, I could not properly understand the second part of your comment, starting with "The optimisation functions...". But as far as I have understood, let me comment on that: I will multiply the singular values with each other as I have showed them in my previous comment. As a result, I'll get a single function, not a vector anymore. Therefore the optimization function I guess will accept the single function as an input.

In addition, can you explain a little bit more the idea of using norm? What exactly is the advantage of using norm?

Thanks,

Erdem

[Star Strider]

My pleasure!

This is my idea:

T = rand(6,1)*10;                           % Desired SVD of the matrix
ObjFn = @(kx) norm(T - svd(Erdem(kx)));     % Objective Function for ‘fminsearch’
[kx, fval] = fminsearch(ObjFn, 1)           % Do the fit
Chk = svd(Erdem(kx))                        % Check the result

When I ran it it worked, but didn’t converge on the actual value of T. When I plotted it, I couldn’t find a value of kx that resulted in a zero-crossing, so you will likely have to experiment with it to get it to produce the result you want, including complex-valued kx. (If you have access to the Optimization Toolbox, you may want to see if other functions give better results.) You can also experiment with optimset to tighten the tolerances on fminsearch and otherwise alter its behaviour. (If the optimisation is successful, fval should be close to zero.) I’ve never attempted to do curve-fitting (essentially what this is) with the svd function, so this is a first for us all.

EDIT — Forgot to mention norm. The norm function calculates the euclidean distance between T and the result of the svd. It returns a scalar, which is what fminsearch wants. I’ve used it before to ‘fit’ vector-valued functions in this sort of situation. (To give credit where credit is due, Matt J enlightened me to its use in these situations.)

[Erdem]

Thank you so much. I achieved to obtain a converging solution. There are some problems in the solution, but these can be discussed under a different thread. I'll open a different thread for these.

Erdem

[Star Strider]

My pleasure!

I’m very happy it works! I’ll look for your other thread, since I’m interested in your project.