I have the following expression to implement in MATLAB:
(A*u).*(B*v)
Where A and B are matrices N x N and u and v column vectors N x 1.
For some reason, not really important at the moment, I want to represent it in the form C(A,u,b)*v for some matrix N x N originated by A,u and B.
How can I do that?
I tried to do something like using the following matrix
C = spdiags( A*u,0,N,N)*B
But it seems that the result is not really what I need.
How about
C1 = bsxfun(#times, A*u,B) * v
Related
Suppose we have a Mx3 matrix and a 1x3 vector. How can I compute the dot product of each column and the vector without using a loop?
If I understood your question correctly,
M = rand(8,3); V = 1:3;
P = sum( M .* V, 2 ); % or in older MATLAB versions: sum( bsxfun(#times, M, V), 2 );
If you're dealing with complex numbers, you might have to conjugate one of the inputs.
Using Dev-iL’s example data:
M = rand(8,3);
V = 1:3;
the dot product of each row of M with V is simply the matrix product with a transposed V:
M * V'
Note that ' returns the conjugate transpose, which you need for the dot product.
Computing the dot product with the columns of M, as stated in the question, is nonsensical because the dimensions don’t match, hence I presume you meant rows (as the other answers did).
For a 3N by 3N by 3N matrix A, I would like to derive a N by N by N matrix B whose entries come from summation over blocks in A.
For example, B(1,1,1) = sum of all elements of A(1:3,1:3,1:3).
Basically, A is kind of a high resolution matrix and B is a low resolution matrix from summing over entries in A.
If memory is not a concern, you can use a "labelling" approach: build a 3-component label to group the elements of A, and use that label as the first input argument to accumarray to do the sum. The label uses integers from 1 to N, so the result of accumarray already has the desired shape (NxNxN).
N = 5;
F = 3; %// block size per dimension
A = rand(15,15,15); %// example data. Size FN x FN x FN
[ii jj kk] = ind2sub(size(A), 1:numel(A));
label = ceil([ii.' jj.' kk.']/F);
result = accumarray(label, A(:));
reshape + sum based approach and as such has to be pretty efficient -
sumrows = sum(reshape(A,3,[]),1); %// Sum along rows
sumcols = sum(reshape(sumrows,N,3,[]),2); %// Sum along cols
B = reshape(sum(reshape(sumcols,N*N,3,[]),2),N,N,N); %// Sum along 3rd dim
If you are crazy about one-liners, here's that combining all steps into one -
B = reshape(sum(reshape(sum(reshape(sum(reshape(A,3,[]),1),N,3,[]),2),N*N,3,[]),2),N,N,N);
For a 2D matrix, this would work:
B = reshape(sum(im2col(A, [3 3], 'distinct')), [N N]);
NB: You need the image processing toolbox.
But for 3D matrices, I don't know of any built-in function equivalent to im2col. You might have to use a loop. Left as an exercise to the reader ;)
For a specific problem, I need to design the resizing of a matrix process using multiplication of matrices alone.
Given a matrix of A of dimensions (a*b,1) where a and b are integers, I need to find a way to resize A to dimensions (a,b) like this:
M*A*N = resize(A,a,b)
where dim(M) = (a,a*b) and dim(N) = (1,b). It doesn't have to be two matrices but I don't think it is possible any other way.
If you can't use reshape or vec2mat, you need to do your manipulation for each element of A separately.
There is no such M and N that you are searching for.
Suppose:
resh_A = M*A*N;
Let's study one row of this equation. Assume one row of M*A :
temp_i = M(i, :) * A;
Since M(i, :) is 1 x a*b and A is a*b x 1; temp_i whould be a 1 x 1 matrix.
Now temp_i * N should result in the ith row of your result (or resh_A).
Thus resh_A will look like:
(note N is 1 x b)
temp_1 * N % row1
temp_2 * N % row2
temp_3 * N % row3
...
which is not a general matrix (it's a matrix with rank 1).
I have a simple function below (I omitted the allocations, etc. for brevity) that I have been tryig to plot against it's x value for specific values of N and T but I keep getting a dimensions error. I think that when I try to plot this I am defining an array for x and then plotting Psum(N', x, T') for certain N' and T' against these x, however MATLAB doesn't seem to like this. Can someone give me some direction please.
function U = Psum(N, X, T)
for m = 1:N
A(1,m) = (1/(m*pi))*sin(m*pi*X)*T*exp(-(m^2)*(pi^2)*T);
% array terms of partial sum
end
M = -sum(A); % evaluate Nth partial sum
U = T*(1-X) + M; % output U(X,T) = T(1-X) + V(X,T)
end
I'm getting a similar error when I try to plot the following, I think there must be something wrong with my general approach
syms x;
f = #(x)((x/(100*pi))*(exp(-(100^2)*(pi^2)*x)));
x = 0:0.1:10000;
plot(x,f(x),'r')
title('PartialSum convergence');
xlabel('T');
ylabel('a_n');
the error I get here reads:
Error using *
Inner matrix dimensions must agree.
Here's the analysis of why you're getting a dimension mismatch error. From this line:
A(1,m) = (1/(m*pi))*sin(m*pi*X)*T*exp(-(m^2)*(pi^2)*T)
The element A(1, m) is supposed to be a scalar value in a two-dimensional matrix. Now let's see what are the dimensions of each of the multiplicands:
(1/(m*pi)) is a scalar (that is, a 1×1 matrix).
sin(m*pi*X) has the same dimensions as X. Let's assume its dimensions are q×n.
exp(-(m^2)*(pi^2)*T) has the same dimensions as T, and is multiplied by T.
Therefore T must be a square matrix, so let's assume its dimensions are p×p.
What we get is a q×n matrix multiplied by a square p×p matrix, and the result must be a scalar (that is, 1×1 matrix). This forces q=1 and n=p.
Now let's look at this line:
U = T*(1-X) + M
We are forced to conclude that p=1, otherwise T cannot be multiplied by X from the right.
This means that your code forces T and X to be scalar! No wonder you're getting a error :)
The remedy is simple: revise the computation in Psum so that it can produce correct results for both a scalar X and a vector X. A possible fix would be adding another loop to iterate over all values of X:
function U = Psum(N, X, T)
U = zeros(size(X));
for k = 1:numel(X) %// Iterate over all values of X
for m = 1:N
A(1,m) = (1/(m*pi))*sin(m*pi*X(k))*T*exp(-(m^2)*(pi^2)*T);
%// array terms of partial sum
end
M = -sum(A); % evaluate Nth partial sum
U(k) = T*(1-X(k)) + M; % output U(X,T) = T(1-X) + V(X,T)
end
end
The output of this function has the same dimensions as X.
By the way, did you verify that Psum produces that correct result for scalar inputs?
I don't fully understand what you are trying to accomplish, but just an observation for you: if your input X is a vector, line 3 can not be computed correctly
A(1,m) = (1/(m*pi))*sin(m*pi*X)*T*exp(-(m^2)*(pi^2)*T);
because the right hand side of the equation give you a vector, but the right hand side A(1,m) is one element, not vector. so you have dimension mismatch.
Hope this helps!
I have got a problem like A*x=lambda*x, where A is of order d*d, x is of order d*c and lambda is a constant. A and lambda are known and the matrix x is unknown.
Is there any way to solve this problem in matlab?? (Like eigen values but x is a d*c matrix instead of being a vector).
If I've understood you correctly, there will not necessarily be any solutions for x. If A*x=lambda*x, then any column y of x satisfies A*y=lambda*y, so the columns of x are simply eigenvectors of A corresponding to the eigenvalue lambda, and there will only be any solutions if lambda is in fact an eigenvalue.
From the documentation:
[V,D] = eig(A) produces matrices of eigenvalues (D) and eigenvectors
(V) of matrix A, so that A*V = V*D. Matrix D is the canonical form of
A — a diagonal matrix with A's eigenvalues on the main diagonal.
Matrix V is the modal matrix — its columns are the eigenvectors of A.
You can use this to check if lambda is an eigenvalue, and find any corresponding eigenvectors.
You can transform this problem. Write x as vector by by using x(:) (has size d*c x 1). Then A can be rewritten to a d*c x d*c matrix which has c versions of A along the diagonal.
Now it's a simple eigenvalue problem.
Its actually trivial. Your requirement is that A*X = lambda*X, where X is an array. Effectively, look at what happens for a single column of X. If An array X exists, then it is true that
A*X(:,i) = lambda*X(:,i)
And this must be true for the SAME value of lambda for all columns of X. Essentially, this means that X(:,i) is an eigenvector of A, with corresponding eigenvalue lambda. More importantly, it means that EVERY column of X has the same eigenvalue as every other column.
So a trivial solution to this problem is to simply have a matrix X with identical columns, as long as that column is an eigenvector of A. If an eigenvalue has multiplicity greater than one (therefore there are multiple eigenvectors with the same eigenvalue) then the columns of X may be any linear combination of those eigenvectors.
Try it in practice. I'll pick some simple matrix A.
>> A = [2 3;3 2];
>> [V,D] = eig(A)
V =
-0.70711 0.70711
0.70711 0.70711
D =
-1 0
0 5
The second column of V is an eigenvector, with eigenvalue of 5. We can arbitrarily scale an eigenvector by any constant. So now pick the vector vec, and create a matrix with replicated columns.
>> vec = [1;1];
>> A*[vec,vec,vec]
ans =
5 5 5
5 5 5
This should surprise nobody.