Problem about RootOf in Singular Value Decomposition

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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

Accepted Answer

Star Strider
Star Strider on 12 Jul 2014
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.
  9 Comments
Erdem
Erdem on 13 Jul 2014
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
Star Strider on 13 Jul 2014
My pleasure!
I’m very happy it works! I’ll look for your other thread, since I’m interested in your project.

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