Solving a matrix in MATLAB? - math

How does one solve the (non-trivial) solution Ax = 0 for x in MATLAB?
A = matrix
x = matrix trying to solve for
I've tried solve('A * x = 0', 'x') but I only get 0 for an answer.

Please note that null(A) does the same thing (for a rank-deficient matrix) as the following, but this is using the svd(A) function in MATLAB (which as I've mentioned in my comments is what null(A) does).
[U S V] = svd(A);
x = V(:,end)
For more about this, here's an link related to this (can't post it to here due to the formulae).
If you want a more intuitive feel of singular and eigenvalue decompositions check out eigshow in MATLAB.

You can use N = null(A) to get a matrix N. Any of the columns of N (or, indeed, any linear combination of columns of N) will satisfy Ax = 0. This describes all possible such x - you've just found an orthogonal basis for the nullspace of A.
Note: you can only find such an x if A has non-trivial nullspace. This will occur if rank(A) < #cols of A.

You can see if MATLAB has a singular value decomposition in its toolbox. That will give you the null space of the vector.

null(A) will give you the direct answer. If you need a nontrivial solution, try reduced row echelon form and refer the first page of the pdf.
R = rref(A)
http://www.math.colostate.edu/~gerhard/M345/CHP/ch7_4.pdf

Related

Questions about SVD, Singular Value Decomposition

I am not a mathematician, so I need to understand what SVD does and WHY more than how it works exactly from the math perspective. (I understand at least what is the decomposition though).
This guy on youtube gave the only human explanation of SVD saying, that the U matrix maps "user to concept correlation" Sigma matrix defines the strength of each concept, and V maps "movie to concept correlation" given that initial matrix M has users in the rows, and movie (ratings) in the columns.
He also mentioned two concept specifically "sci fi" and "romance" movies. See the picture below.
My questions are:
How SVD knows the number of concepts. He as human mentioned two - sci fi, and romance, but in reality in resulting matrices are 3 concepts. (for example matrix U - that one with blue titles - has 3 columns not 2).
How SVD knows what is the concept after all. I mean, what If i shuffle the columns randomly how SVD then knows what is sci fi, what is romance. I mean, I suppose there is no rule, group the concepts together in the column order. What if scifi movie is the first and last one? and not first 3 columns in the initial matrix M?
What is the practical usage of either U, Sigma or V matrices? (Except that you can multiply them to get the initial matrix M)
Is there also any other possible human explanation of SVD than the guy up provided, or it is the only one possible function? Matrices of correlations.
As was pointed out in the comments you may well get better explanations elsewhere. However since the question is still open, here is my tuppence worth.
Throughout I'll suppose that A is mxn where m>=n, ie that A has more rows than columns.
First of all there are many forms of the SVD, differing in the sizes of the matrices. They all share the fundamental properties that
A = U*S*V'
S is diagonal
U and V have orthogonal columns (ie U'*U = I, V'*V = I)
Perhaps the most useful from a theoretical point of view is the 'full fat' svd where we have that U is mxm, S is mxn and V is nxn. However this has rather a lot of elements that don't really contribute to A. For example S being diagonal we can write
S = ( S1 ) (where S1 is nxn )
( 0 )
If we divide up U into
U = ( U1 U2) (where U1 is mxn and U2 is (mx(m-n)))
Then its straightforward to calculate that
U*S = U1*S1
and so we can throw away the last m-n columns of U and the last m-n rows of S, and still recover A.
Moreover some of the diagonal elements of S1 may be 0; suppose in fact that p<n of them are non zero. Then we can write
S1 = ( S2 0)
( 0 0)
And arguing as above for U and analogously for V' we can in fact throw away all but the first p columns of U and all of S but S2, and all but the first p rows of V, and still recover A.
This latter is the form of SVD ('thin') in your question:
U is mxp
S is pxp
V' is pxm
where p is the number of non-zero singular values of A. This is my answer to your 1.
By convention the elements of S decrease as you move down the diagonal. To achieve this the routine that calculates the svd in effect works with a version of A with shuffled columns. This shuffling is undone by incorporating the shuffle in the U and V' output. This is my answer to your 2: however you shuffle A, it will be in effect shuffled again to ensure that the singular values decrease down the diagonal.
I struggle to answer 3, because I suspect that our ideas of 'practical' are rather different.
One thing that I think practical is to find simpler approximations to A. The reconstruction of A can be written
A = Sum{ 1<=i<=p | U[i]*S[i]*V[i]' }
where the S[i] are the diagonal elements of S, U[i] are the columns of U and V[i] those of V
We might want to use a simpler model for A, for example want to simplify it down to just one term. That is, we might wonder how much we would lose by using fewer 'concepts'. The 'thin' svd above has already done this in the sense that it has thrown away all the coluns that make no contribution to A. In an extreme case, we might wonder what we would get if we reduced to just one concept. This approximation is found by taking just the first term of the sum above. This extends to however many terms -- q say -- we want to allow: we just take the first q terms of the sum above.
I'm sorry, I can't answer 4.

How to select n objects from a set of N objects, maximizing the sum of pairwise distances between them

You have a set of N=400 objects, each having its own coordinates in a, say, 19-dimensional space.
You calculate the (Euclidean) distance matrix (all pairwise distances).
Now you want to select n=50 objects, such that the sum of all pairwise distances between the selected objects is maximal.
I devised a way to solve this by linear programming (code below, for a smaller example), but it seems inefficient to me, because I am using N*(N-1)/2 binary variables, corresponding to all the non-redundant elements of the distance matrix, and then a lot of constraints to ensure self-consistency of the solution vector.
I suspect there must be a simpler approach, where only N variables are used, but I can't immediately think of one.
This post briefly mentions some 'Bron–Kerbosch' algorithm, which apparently addresses the distance sum part.
But in that example the sum of distances is a specific number, so I don't see a direct application to my case.
I had a brief look at quadratic programming, but again I could not see the immediate parallel with my case, although the 'b %*% bT' matrix, where b is the (column) binary solution vector, could in theory be used to multiply the distance matrix, etc.; but I'm really not familiar with this technique.
Could anyone please advise (/point me to other posts explaining) if and how this kind of problem can be solved by linear programming using only N binary variables?
Or provide any other advice on how to tackle the problem more efficiently?
Thanks!
PS: here's the code I referred to above.
require(Matrix)
#distmat defined manually for this example as a sparseMatrix
distmat <- sparseMatrix(i=c(rep(1,4),rep(2,3),rep(3,2),rep(4,1)),j=c(2:5,3:5,4:5,5:5),x=c(0.3,0.2,0.9,0.5,0.1,0.8,0.75,0.6,0.6,0.15))
N = 5
n = 3
distmat_summary <- summary(distmat)
distmat_summary["ID"] <- 1:NROW(distmat_summary)
i.mat <- xtabs(~i+ID,distmat_summary,sparse=T)
j.mat <- xtabs(~j+ID,distmat_summary,sparse=T)
ij.mat <- rbind(i.mat,"5"=rep(0,10))+rbind("1"=rep(0,10),j.mat)
ij.mat.rowSums <- rowSums(ij.mat)
ij.diag.mat <- .sparseDiagonal(n=length(ij.mat.rowSums),-ij.mat.rowSums)
colnames(ij.diag.mat) <- dimnames(ij.mat)[[1]]
mat <- rbind(cbind(ij.mat,ij.diag.mat),cbind(ij.mat,ij.diag.mat),c(rep(0,NCOL(ij.mat)),rep(1,NROW(ij.mat)) ))
dir <- c(rep("<=",NROW(ij.mat)),rep(">=",NROW(ij.mat)),"==")
rhs <- c(rep(0,NROW(ij.mat)),1-unname(ij.mat.rowSums),n)
obj <- xtabs(x~ID,distmat_summary)
obj <- c(obj,setNames(rep(0, NROW(ij.mat)), dimnames(ij.mat)[[1]]))
if (length(find.package(package="Rsymphony",quiet=TRUE))==0) install.packages("Rsymphony")
require(Rsymphony)
LP.sol <- Rsymphony_solve_LP(obj,mat,dir,rhs,types="B",max=TRUE)
items.sol <- (names(obj)[(1+NCOL(ij.mat)):(NCOL(ij.mat)+NROW(ij.mat))])[as.logical(LP.sol$solution[(1+NCOL(ij.mat)):(NCOL(ij.mat)+NROW(ij.mat))])]
items.sol
ID.sol <- names(obj)[1:NCOL(ij.mat)][as.logical(LP.sol$solution[1:NCOL(ij.mat)])]
as.data.frame(distmat_summary[distmat_summary$ID %in% ID.sol,])
This problem is called the p-dispersion-sum problem. It can be formulated using N binary variables, but using quadratic terms. As far as I know, it is not possible to formulate it with only N binary variables in a linear program.
This paper by Pisinger gives the quadratic formulation and discusses bounds and a branch-and-bound algorithm.
Hope this helps.

How can I write this line of code in MATLAB (currently R)?

How can I write this line of code in MATLAB (currently R)?
vcov_beta_hat <- c(sigma2_hat) * solve(t(X) %*% X)
My attempt is,
vcov_beta_hat = [sigma2_hat.*((X'*X))];
However I am struggling on what the 'c' is doing in the r code?
Whilst the above answer addresses that the solve is the something missing in your matlab code, solve can mean a number of different things in R,
If there is no comma in the equation its not solving anything and is actually taking the inverse,
Inverse of A, MATLAB: inv(A) R: solve(A)
Therefore, vcov_beta_hat = [sigma2_hat.*inv((X'*X))];
The c(a,b,c) denote a vector in R. In Matlab, you would write
vec = [a b c];
Also, you need to find the equivalent of the R-solve() function. So far, your matlab code just mutliplies X' with X and does not solve the system of equations.
linsolve should be a good starting point.

How to solve multivariable nonlinear equotation systems? [R]

I have three equotations with 3 unknown variables like this:
Assume that the following variables are given as parameters:
Desired, not given:
What I want is that when I pass three parameters it should give me the solution of the remaining ones. What's the simpliest way to do this in R?
Because I'm a beginner I would like to have some (short) explanations. :)
Solving for sigma.eps^2 is a matter of mathematics. In the second equation you can substitute a (take the expression from the third equation). Then you can solve for sigma.eps^2. After that you can calculate a and then b:
sigma.eps2 <- (1-p^2)*sigma2^2 # sigma.eps2 stands for sigma.eps^2
sigma.eps <- sqrt(sigma.eps2)
a <- (sigma.eps / sigma1) * p / sqrt(1-p^2)
b <- mu2 - a*mu1
Eventually the second value for sigma.eps is relevant. In this case the second value is:
sigma.eps <- -sqrt(sigma.eps2)
This would also implicate other values for a and b (to compute in the same way as above).

troubles with integration on matlab

I'd like some help please I really need to solve this problem.
Well before anything thank you for your time...
My problem: I have a matrix (826x826 double) and I want to integrate this matrix with respect to a vector of (826x1 double) I don't have the functions of any of this. Is there a command or an algorithm to take the integral of a matrix with respect to a vector? Please I really need help, I'm such a newbie at matlab.
Sincerely.
George
If it's a constant matrix A integrated with respect to vector x, your answer in simply Ax + c where c is some constant vector. If A is a function of x, you will need to specify exactly what it is. Another case is when both A and x are functions of t. There is no one simple answer and no computer program would do it in most cases. There are books written in this stuff. It's not an easy task.
If I understand correctly, you have a matrix Y (size mxn) and a vector X (size mx1) where Y(i, j) = f_j(X(i)) for some unknown function f_j. To approximate the integral of each column over X you could use the trapz function of Matlab which uses the trapezoidal method.
A = trapz(X, Y);
This will integrate Y along its columns using the vector X. If you wanted to integrate along rows you can call the trapz function with an added argument of dim=2. Of course, the dimensions of X and Y must be compatible in either case.

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