I need to understand a part of the GMRES algorithm written in these languages. What makes me a problem is the following piece:
y = H(1:k,1:k) \ beta(1:k);
x = x + Q(:,1:k)*y;
Can someone explain what does it mean in extended form.
Thank you in advance.
For what concerns the first equation:
H(1:k,1:k) = sub-matrix of matrix H that you obtain by taking rows from 1 (beginning) to k and columns from 1 (beginning) to k
beta(1:k) = sub-vector of vector beta that you obtain by taking elements from 1 (beginning) to k
y = is a matrix obtained by solving a symbolic matrix left division between sub-matrix of H and the sub-vector of beta
For what concerns the second equation:
Q(:,1:k) = sub-matrix of matrix Q with all the rows and columns from 1 (beginning) to k
x = a matrix that is obtained by adding to it's previous value the result of the multiplication between the sub-matrix of matrix Q and y
Indexing in Matlab is 1-based, not 0-based. So index 1 corresponds to the first element of whatever you are working with. Example of sub-matrix by indexing:
A = [
2 3 4;
1 2 3;
3 4 4
];
B = A(1:2,1:2);
B is then equal to:
[
2 3;
1 2
];
C = A(:,1:2);
C is then equal to:
[
2 3;
1 2;
3 4
];
That weird division symbol represents a matrix left division (for more information: mathworks.com/help/symbolic/mldivide.html): X = A\B solves the symbolic system of linear equations in matrix form: A*X = B for X.
Related
I have a matrix A = [1 1 2; 2 1 3; 3 1 4; 4 1 5]; with rank of 2. I want to find the orthonormal basis of the column space of this matrix using Singular Value Decomposition (SVD):
[U S V] = svd(A);
Now, since the rank is 2, the first 2 columns of the left singular vector matrix, U should provide me the orthonormal basis vectors for the column space of matrix A. But, using rref(A) and calculating the column space gives me these orthonormal basis vectors:
[0.1825, 0.36514, 0.5477, 0.7303] and [0.5, 0.5, 0.5, 0.5] which is not the same as the first 2 column vectors of U.
How can I obtain the basis vectors using svd()?
a=sym(y);
display(a);
rank(a);
display(rank(a));
at=transpose(a);
nullspace = null(a);
display(nullspace);
%Display the column space,row space and null space of a matrix
a=[1 1 2; 2 1 3; 3 1 4; 4 1 5];
nullspace = null(a);
display(nullspace);
columnspaceA = colspace(a);
display(columnspaceA);
leftnullspace = null(at);
display(leftnullspace);
rowspaceA = colspace(at);
display(rowspaceA);
Lets say I have matrices A = [1 2; 3 4], B = [4 3; 2 1]. I want to multiply each column from matrix A ([1; 3], [2; 4]) by the corresponding row in matrix B ([4 3], [2 1]) and sum resulting matrices. I have came up with the following code:
C = zeros(size(A));
for i = 1 : size(A, 1)
C = C + A(:, i) * B(i, :);
end
Could it be rewritten using some math trick or matlab function to get rid of the for loop?
I see there is unclarity in my question regarding the result I want. The result should exactly mimic provided Matlab code, therefore I seek one matrix which is given by the matrix summation of the intermediate matrices that are created by multiplying each column vector with corresponding row vector from both matrices. For this specific example, it would be given by
C = A(:, 1) * B(1, :) + A(:, 2) * B(2, :);
I am just looking for some generic, for-loop less version for any matrices of compatible dimensions.
I just tried out my suggestion in the comments, and it seems to work with this octave tester:
Short form (only works in Octave):
A = [1 2; 3 4], B = [4 3; 2 1]
X = sum((A * B)(:))
Long form (Matlab):
A = [1 2; 3 4]
B = [4 3; 2 1]
C = A * B % Stop here if you want the exact result from your Matlab code
x = sum(C(:)) % To get the sum of the resulting matrix
Sources:
https://www.tutorialspoint.com/matlab/matlab_matrix_multiplication.htm
https://www.mathworks.com/matlabcentral/newsreader/view_thread/51252
Update, based on your update:
Output of A * B:
8 5
20 13
Output of your code:
8 5
20 13
It appears that
C = zeros(size(A));
for i = 1 : size(A, 1)
C = C + A(:, i) * B(i, :);
end
is equivalent to the matrix multiplication
C = A*B
for square matrices A and B.
You can also do this in MATLAB, to get the sum.
C=ones(1,2)*A*B*ones(2,1)
The general form would be
C=ones(1,size(A,1))*(A*B)*ones(size(B,2),1);
EDIT
I see you updated your question for clarity. The matrix product can be calculated directly
C = A*B;
as pointed out by jodag.
This works provided you follow rules of linear algebra where inner dimensions of your matrices are the same (i.e. when number of columns in A match the number of rows in B; size(A,2)==size(B,1)).
I have a sparse parity check matrix (consisting of ones and zeros) and I would need to replace each nonzero element (ones) from it by PM: a squared permutation matrix of dimension N (being N generally a large integer). In the case of the zero elements, these would be replaced by squared null matrices of the same dimension.
Let me share with you which the current state of my code is:
This is the base matrix in which I would like to replace its ones by permutation matrices:
B = zeros((L + ms) * dc, L * dv);
for i = 1 : 1 : L
for j = 1 : 1 : dv
B(dc*(i-1)+1 : dc*(ms+i), j+dv*(i-1)) = ones(dc*(ms+1), 1);
end
end
I have been told a way for doing so by using 'cell' objects, this is, initializing H as an array of empty cells that would contain the corresponding submatrices:
H=repmat({{}},size(B));
Mc = 500 % Dimension of the permutation matrix
MP=randi([1,5],Mc,Mc); % Definition of one permutation matrix
% It would be desirable that the permutation matrix is different for each replacement
[H{B==0}]=deal(zeros(Mp));
[H{B==1}]=deal(MP);
But there's one problem coming up -that I would need this matrix to be used as a parameter of a following function and it would be very much desirable that it were a simple matrix of ones and zeros (as I am not very familiar with 'cell' structures... however, as you can see, Mc is such a big integer that I don't know if that would be possible to be handled.
Do you have any other way of doing so to have a raw matrix of dimensions (L*ms)dcMc, LdvMc as output?
These are some parameters that could be used to have a try:
ms = 2;
Mc = 600; % any number (specially big ones) could serve for this purpose
dc = 3;
dv = 4;
L = 15;
Many thanks in advance for your attention, and may you have a nice day.
This is how it can be done with cells:
B = Randi([0,1], m, n); % m*n array of 1s and 0s
H = cell(size(B)); % Define cell array
Mc = 500; % Size of replacement matrices
MP = randi([1,5], Mc, Mc); % Permutation matrix
H(B==0) = {zeros(Mc, Mc)}; % Set elements of H where B==0 to matrix of zeros
H(B==1) = {MP}; % Set elements of H where B==1 to permutation matrix
% Convert to matrix
Hmat = cell2mat(H); % Hmat = Mc*m row by Mc*n column matrix
Each cell element holds a matrix of size Mc*Mc, at the end this can be converted to one large matrix. Take care which sorts of brackets you use with cells, the parentheses () are for logical indexing, whilst the curly braces {} are for assigning the sub-matrixes as cell elements.
Example:
B = [1 0; 1 1];
MP = [1 2; 3 4];
H(B==0) = {zeros(2, 2)};
H(B==1) = {MP};
Hmat = cell2mat(H);
% >> Hmat = [1 2 0 0
% 3 4 0 0
% 1 2 1 2
% 3 4 3 4];
If you want the replacement matrix MP to change, you will have to do this in a loop, changing MP on each iteration and using it to replace one element of H.
I have two vectors, X of bases and N of exponents. I want to get the matrix of all values e = xn for each x in X and n in N.
For example, the following input:
X = [2 3 4]'
N = [1 2 3]
should produce:
ans = [2 4 8; 3 9 27; 4 16 64]
Is there a way to get this without looping (just like you can get all values of x×n by using the column by row product)?
Use bsxfun:
bsxfun(#power, X, N)
This assumes that X is a column vector and N is a row vector. If you want to guarantee that, use the following syntax which is more robust:
bsxfun(#power, X(:), N(:).')
This is probably a bit sloppier than the bsxfun answer, but you could use meshgrid:
E = X.^(meshgrid(N)')
This assumes both X and N are row vectors. If both are column vectors then it becomes:
E = X.^(meshgrid(N))
I have two vectors, X of bases and N of exponents. I want to get the matrix of all values e = xn for each x in X and n in N.
For example, the following input:
X = [2 3 4]'
N = [1 2 3]
should produce:
ans = [2 4 8; 3 9 27; 4 16 64]
Is there a way to get this without looping (just like you can get all values of x×n by using the column by row product)?
Use bsxfun:
bsxfun(#power, X, N)
This assumes that X is a column vector and N is a row vector. If you want to guarantee that, use the following syntax which is more robust:
bsxfun(#power, X(:), N(:).')
This is probably a bit sloppier than the bsxfun answer, but you could use meshgrid:
E = X.^(meshgrid(N)')
This assumes both X and N are row vectors. If both are column vectors then it becomes:
E = X.^(meshgrid(N))