Suppose I have the matrix below:
syms x y z
M = [x+y-z;2*x+3*y+-5*z;-x-y6*z];
I want to have the a matrix consisting the coefficients of the variables x,y, and z:
CM = [1,1,-1;2,3,-5;-1,-1,6];
If I multiply CM by [x;y;z], I expect to get M.
Edit
I have a system of ODE:
(d/dt)A = B
A and B are square matrices. I want to solve these set of equations. I don't want to use ode solving commands of Matlab.
If I turn the above set of equations into:
(d/dt)a = M*a
then I can solve it easily by the eigen vectors and values of matrix M. Here a is a column vector containing the variables, and M is the matrix of coefficient extracted from B.
Since you seem to be using the Symbolic Math Toolbox, you should diff symbolically, saving the derivative with respect to each variable:
syms x y z;
M=[x+y-z;2*x+3*y-5*z;-x-y+6*z];
Mdiff=[];
for k=symvar(M)
Mdiff=[Mdiff diff(M,k)];
end
Then you get
Mdiff =
[ 1, 1, -1]
[ 2, 3, -5]
[ -1, -1, 6]
If you want to order the columns in a non-lexicographical way, then you need to use a vector of your own instead of symvar.
Update
Since you mentioned that this approach is slow, it might be faster to use coeffs to treat M as a polynomial of its variables:
syms x y z;
M=[x+y-z;2*x+3*y-5*z;-x-y+6*z];
Mdiff2=[];
varnames=symvar(M);
for k=1:length(M)
Mdiff2=[Mdiff2; coeffs(M(k),varnames(end:-1:1))];
end
Note that for some reason (which I don't understand) the output of coeffs is reversed compare to its input variable list, this is why we call it with an explicitly reversed version of symvar(M).
Output:
>> Mdiff2
Mdiff2 =
[ 1, 1, -1]
[ 2, 3, -5]
[ -1, -1, 6]
As #horchler pointed out, this second solution will not work if your symbolic vector has varying number of variables in its components. Since speed only matters if you have to do this operation a lot of times, with many configurations of the parameters in your M, I would suggest constructing M parametrically (so that the coefficients are also syms) is possible, then you only have to perform the first version once. The rest is only substitution into the result.
Related
This problem was discovered when I was solving this second order differential equation by matlab
syms x(t)
diff(x,t,2)+k*diff(x,t)+b*x==0
I transform it into a system of first order equations, whose coefficient matrix is
syms b k
A=[0 1;
-b-k]
Then I solve this system by following code
syms x1(t) x2(t)
X=[x1;x2];
odes=diff(X)==A*X;
[x1sol(t),x2sol(t)]=dsolve(odes);
x1sol=simplify(x1sol(t))
x2sol=simplify(x2sol(t))
But this solution is different from the one I calculated manually, because the correct answer is based on the eigenvalues and eigenvectors of A:
λ1=(-b-(b^2-4*k)^(1/2))/2
v1=[-1;
(b+(b^2-4*k)^(1/2))/2]
and
λ2=(-b+(b^2-4*k)^(1/2))/2
v2=[-1;
(-b+(b^2-4*k)^(1/2))/2]
Therefore, Instead of solving the system directly, I calculate the the eigenvalues and eigenvectors of A directly
[V,D]=eig(A)
or indirectly
syms k b t
I=eye(2);
A=[0 1;-k -b];
e=eig(A);
B1=e(1)*I-A;
B2=e(2)*I-A;
P1=null(B1)
P2=null(B2)
By comparing the results of MATLAB and manual calculation, I find that the eigenvalues are correct calculated by both means, but the eigenvectors are wrong calculated by both means either (and they are same with dsolve(odes) above). If P1 is the solution to B1X=0 (means P1 is one of the eigenvector of A), then B1P1=0 should be true, so does P2.
Finally, I find that MATLAB seems to make an error when converting B1 a to the simplest row form
B1=
[- b/2 - (b^2 - 4*k)^(1/2)/2, -1]
[ k, b/2 - (b^2 - 4*k)^(1/2)/2]
rrefB1=rref(B1)
ans=
[1, (b - (b^2 - 4*k)^(1/2))/(2*k)]
[0, 0]
and I check rrefB1 multiplies P1, and it equals to 0.
So, the problem is that B1 multiplies P1 doesn't equals to 0, but rrefB1 multiplies P1 equals to 0.
In theory, The original matrix (B1) and the simplest row form of it (rrefB1) should have the same fundamental system of solutions.
What's wrong here?
I am trying to solve a second order differential using ODE45 in Matlab with matrix as inputs. I am struck with couple of errors that includes :
"In an assignment A(I) = B, the number of elements in B and
I must be the same."
Double order differential equations given below:
dy(1)= diag(ones(1,100) - 0.5*y(2))*Co;
dy(2)= -1 * Laplacian(y(1)) * y(2);
Main function call is:
[T,Y] = ode45(#rigid,[0.000 100.000],[Co Xo]);
Here, Co is Matrix of size 100x100 and Xo is a column matrix of size 100x1. Laplacian is a pre-defined function to compute matrix laplacian.
I will appreciate any help in this. Should I reshape input matrices and vectors to fall in same dimensions or something?
Your guess is correct. The MATLAB ode suite can solve only vector valued ode, i.e. an ode of the form y'=f(t,y). In your case you should convert y, and dy, back and forth between a matrix and an array by using reshape.
To be more precise, the initial condition will be transformed into the array
y0 = reshape([Co Xo], 100*101, 1);
while y will be obtained with
y_matrix = reshape(y, 100, 101);
y1 = y_matrix(:,1:100);
y2 = y_matrix(:,101);
After having computed the matrices dy1 and dy2 you will have to covert them in an array with
dy = reshape([dy1 dy2], 100*101, 1);
Aside from the limitations of ode45 your code gives that error because, in MATLAB, matrices are not indexed in that way. In fact, if you define A = magic(5), A(11) gives the eleventh element of A i.e. 1.
I'm trying to figure out Eigenvalues/Eigenvectors for large datasets in order to compute
the PCA. I can calculate the Eigenvalues and Eigenvectors for 2x2, 3x3 etc..
The problem is, I have a dataset containing 451x128 I compute the covariance matrix which
gives me 128x128 values from this. This, therefore looks like the following:
A = [ [1, 2, 3,
2, 3, 1,
..........,
= 128]
[5, 4, 1,
3, 2, 1,
2, 1, 2,
..........
= 128]
.......,
128]
Computing the Eigenvalues and vectors for a 128x128 vector seems really difficult and
would take a lot of computing power. However, if I allow for each of the blocks in A to be a 2-dimensional (3xN) I can then compute the covariance matrix which will give me a 3x3 matrix.
My question is this: Would this be a good or reasonable assumption for solving the eigenvalues and vectors? Something like this:
A is a 2-dimensional vector containing 128x451,
foreach of the blocks compute the eigenvalues and eigenvectors of the covariance vector,
like so:
Eig1 = eig(cov(A[0]))
Eig2 = eig(cov(A[1]))
This would then give me 128 Eigenvalues (for each of the blocks inside the 128x128 vector)..
If this is not correct, how does MATLAB handle such large dimensional data?
Have you tried svd()
Do the singular value decomposition
[U,S,V] = svd(X)
U and V are orthogonal matrices and S contains the eigen values. Sort U and V in descending order based on S.
As kkuilla mentions, you can use the SVD of the original matrix, as the SVD of a matrix is related to the Eigenvalues and Eigenvectors of the covariance matrix as I demonstrate in the following example:
A = [1 2 3; 6 5 4]; % A rectangular matrix
X = A*A'; % The covariance matrix of A
[V, D] = eig(X); % Get the eigenvectors and eigenvalues of the covariance matrix
[U,S,W] = svd(A); % Get the singular values of the original matrix
V is a matrix containing the eigenvectors, and D contains the eigenvalues. Now, the relationship:
SST ~ D
U ~ V
As to your own assumption, I may be misreading it, but I think it is false. I can't see why the Eigenvalues of the blocks would relate to the Eigenvalues of the matrix as a whole; they wouldn't correspond to the same Eigenvectors, as the dimensionality of the Eigenvectors wouldn't match. I think your covariances would be different too, but then I'm not completely clear on how you are creating these blocks.
As to how Matlab does it, it does use some tricks. Perhaps the link below might be informative (though it might be a little old). I believe they use (or used) LAPACK and a QZ factorisation to obtain intermediate values.
https://au.mathworks.com/company/newsletters/articles/matlab-incorporates-lapack.html
Use the word
[Eigenvectors, Eigenvalues] = eig(Matrix)
I would like to compute the weighted maxima of a vector in Matlab. For weighted maxima I intend the following:
Given a vector of 2*N+1 weights W={w[-N], w[-N+1] .. w[0] .. w[N]} and given an input sequence A, weighted maxima is a vector M where m[i]=max(w[-N]*a[i-N], w[-N+1]*a[i-N+1], ... w[N]*a[i+N])
So for example given a vector A= [1, 4, 12, 2, 4] and weights W=[0.5, 1, 0.5], the weighted maxima would be M=[2, 6, 12, 6, 4].
This can be done using ordfilt2, but ordfilt2 uses weights as additive rather then multiplicative.
I am actually working on 4D matrixes, but any 1D solution would work as the 4D weight matrix is separable.
My current solution is to generate shifted copies of the input array A, weight them according to the shift and maximize all the arrays. Shift is performed using circshift and is the bottleneck in the process. generating shifted matrixes "manually" trough indexing turned out to be even slower.
Can you suggest any more efficient solution?
EDIT: For a positive A, M=exp(ordfilt2(log(A), length(W), ones(size(W)), log(W))) does the job, but still takes longer than the circshift solution above. I am still looking for more efficient solutions.
>> B = padarray(A, [0 floor(numel(W)/2)], 0); % Pad A with zeros
>> B = bsxfun(#times, B(bsxfun(#plus, 1:numel(B)-numel(W)+1, (0:numel(W)-1)')), W(:)); % Apply the weights
>> M = max(B) % Compute the local maxima
M =
2 6 12 6 4
I want to take a polynomial p as input from user in matlab for a given degree (specified by the user each time) such that the polynomial is input one element at a time into a matrix at each index from 1 to n. where n is the polynomial degree. was tryin to do something like this but m stuck
for M = 1:n
p[n] = input('polynomial')
p
end
How should I input a polynomial coefficient at each index of matrix i.e. how to reach each index position?
Instead of using a loop, you can take a polynomial as input using the following method:
p = input('Enter a polynomial in [] brackets');
Now the user should enter the polynomial like this:
[2, 4, 3, 8];
Then you can calculate its degree using the length command:
n = length(p);