AX=0 and introducing error matrix [closed] - matlab

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I'm having a matrix of 10 rows and 5 columns, thatr I have called A
I would like to make AX=0 and determine X which contains 5 unknown parameters.
What I did is
null(A)
But it seems that it's considering what I did as a linear algebra.
I would like to introduce another matrix which contains a matrix of errors . which would give me a priori:
AX + E = 0
because the results are not accurate, indeed, but still, I would like to find the closest parameters of the unknown vector (X) and get an error matrix.
Can you help me with that?

You can do eigen value decomposition of A^T.
[Q, D] = eigen(A^T)
And take the eigen vectors that corresponds to the smallest eigen value.
That is take y vectors from the right hand side of the Q matrix.
y is the column number of X.
And then E = -AX
====edit=====
OK. actually you want to minimize E. But since E is a matrix I assume you want to minimize the sum of squares of all the elements in E.
That is:
||E||_F^2 = ||-AX||_F^2 = trace(X^T(A^TA)X)
You can do eigen decomposition of A^TA
[Q, D] = eigen(A^TA)
QA^TA = DQ => D = QA^TAQ^T
To make it smaller, X should be the columns of Qs that corresponds to the smallest eigen value in the diagonal of D.
If there is a eigen value equals to 0, then AX can be 0 and E=0

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MATLAB create a matrix M = F(i,j) [closed]

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Suppose I have a function of 2 variables F(i,j) which depends on the row index and column index of a matrix, and I want to fill the matrix with the values M_ij = F(i,j)
Of course it's possible to do a loop through i and j, or even only i or j if the function F can be vectorized, but I'd like to know the neat way to do that.
It's impossible to answer without seeing your F but let's assume that F is vectorized such as
F = #(x,y)x+y;
then you could use ndgrid:
[I,J] = ndgrid(1:m,1:n);
M = F(I,J)
In the above case, and this may well apply in your case as well, you might be able to vectorize the function directly using something like bsxfun:
M = bsxfun(#plus, 0:m-1, 1:n);
Regardless of whether your function F is vectorized or not, you have to evaluate it for every value of i and j. If F is not vectorized, you will have to do a loop over the indices manually. If F is based on MATLAB builtins like sin, log, etc., it is most likely vectorized. In this case, you can pass in i and j that are the same size as M and get the result in one step:
[j, i] = meshgrid(1:size(M, 2), 1:size(M, 3))
M = F(i, j)
Note that meshgrid takes and returns the parameters as X, Y, which is the opposite of matrix indexing order row, col.

Adding Matrices in Matlab with out looping [closed]

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how to sum m-by-n matrix A, to m by n by p matrix B, using Matlab without a For loop. Where the result C should be m by n by p matrix, the direct addition results in
Error using +
Matrix dimensions must agree.
?
If you are looking to add A to each of the p slices of B then you should use bsxfun:
bsxfun(#plus,A,B)
First modify A to have the same size of B by replicating A, p times:
A = repmat(A ,[1 1 p]);
Now A is m by n by p the summation then can be done as
C = A + B
Where C is a m by n by p matrix

Octave beginner help. Creating a function that takes a component matrix as a parameter and returns a matrix? [closed]

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I need to create a function that takes a component matrix as a parameter and returns a matrix?
Apparently this function should normalise my data?
There are other instructions along with this step in my project such as:
Take the matrix and calculate the mean value along a certain column.
Calculate the difference between the measurement and this mean.
Subtract this difference from each measurement.
Return corrected matrix to the script.
Place corrected matrix in a variable within the script.
(I don't know if this is what the function is supposed to do or anything I'm completely lost and any help would be appreciated thanks!)
This is probably homework but I'll help you get started.
To create a function which takes a matrix and return a matrix:
function m_out = my_function(m_in)
%insert calculations here
end
To find the 2-norm of a matrix (which is the largest singular value):
the_norm = norm(my_matrix); % returns a scalar, 2-norm of matrix
To find the mean of a vector:
the_mean = mean(my_vector); % returns a scalar, mean of the vector
To access a specific column of a matrix:
my_col = my_matrix(:, col_number); % my_col is a vector
To access a specific row of a matrix:
my_row = my_matrix(row_num, :); % my_row is a vector
To subtract a scalar (single number) from a matrix:
new_matrix = old_matrix - single_number; % returns a matrix
To store a matrix into a variable (example):
my_matrix = [1,2,3;4,5,6;7,8,9];
Give it a try creating a function which puts this all together.

Creat Log-normal Random Variable in MATLAB [closed]

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I have trouble with the probability density function (p.d.f) of a log-normal distribution. I really need your help. As defined by wikipedia:
http://en.wikipedia.org/wiki/Log-normal_distribution#Probability_density_function
The probability density function of a log-normal distribution is:
My problem is, how to define x variable in MATLAB? Thanks for your help!
If you have the Stats toolbox, you can just use lognpdf:
y = lognpdf(x,mu,sigma);
Though this is a very simple function - fully vectorized, it's effectively just:
y = exp(-0.5*((log(x)-mu)./sigma).^2)./(x.*sqrt(2*pi).*sigma);
But you may want to check that x > 0 and sig > 0. To create this plot on the Wikipedia article that you cited, you can do:
mu = 0;
sigma = [1;0.5;0.25];
x = 0:0.01:3;
y = lognpdf([x;x;x],mu,sigma(:,ones(1,length(x))));
figure; plot(x,y);
When your question asks about defining x, maybe you're actually looking for log-normally distributed random variables, i.e., you want to sample randomly from the log-normal PDF/distribution? In that case you can use lognrnd:
r = lognrnd(mu,sigma);
I'm confused, like you can do this in a one-liner,
fun = #(x,mu,sigma) (1./(x*sigma*sqrt(2*pi))).*exp( -(power((log(x)-mu),2))/(2*power(sigma,2)))
x is any value that satisfies x > 0, the pdf tells you via Wikipedia
In probability theory, a probability density function (pdf), or
density of a continuous random variable, is a function that describes
the relative likelihood for this random variable to take on a given
value.
So any value x given to the log-normal pdf tells you tel relative likelihood that a random variable could be that value.
Consider this toy example:
mu = 1;
sigma = 10;
x = logspace(-2,0,10);
plot( x, fun(x,1,10) )
From this plot as x gets closer to zero, it's relative likelihood of actually taking on that value increases. DISCLAIMER I just threw that function together, it needs to be checked for accuracy, the preceding was for illustration only.

Efficient recalculation of weighted least squares regression when weights change [closed]

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I am performing weighted least squares regression as described on wiki: WLS
I need to solve this equation: $B= (t(X)WX)^{-1}*t(X)Wy$
I use SVD to find: $(t(X)WX)^{-1}$ and store it in a matrix. In addition I store the matrix $H= (t(X)WX)^{-1}*t(X)W$ and simply do the following for any new value of y: B= Hy. This way I am able to save the cost of repeating SVD and matrix multiplications as y's change.
W is a diagonal matrix and generally does not change. However sometimes I change one or two elements on the diagonal in the W matrix. In that case I need to do SVD again and recalc the H matrix. This is clearly slow and time consuming.
My question is: If I know what changed in W and nothing changes in X is there a more efficient method to recalculate (t(X)WX)^-1?
Or put differently is there an efficient analytic method to find B given that only diagonal elements in W can change by a known amount?
There is such a method, in the case that the inverse you compute is a true inverse and not a generalised inverse (ie none of the singular values are 0). However some caution in using this is recommended. If you were doing your sums in infinite precision, all would be well. With finite precision, and particularly with nearly singular problems -- if some of the singular values are very large -- these formulae could result in loss of precision.
I'll call inverse you store C. If you add d (which can be positive or negative) to the m'th weight, then the modified C matrix, C~ say, and the modified H, H~, can be computed like this:
(' denotes transpose, and e_m is row the vector that's all 0, except the m'th slot is 1)
Let
c = the m'th column of H, divided by the original m'th weight
a = m'th row of the data matrix X
f = e_m - a*H
gamma = 1/d + a*c
(so c is a column vector, while a and f are row vectors)
Then
C~ = C - c*c'/gamma
H~ = H + c*f/gamma
If you want to find the new B, B~ say, for a given y, it can be calculated via:
r = y[m] - a*B
B~ = B + (r/gamma) * c
The derivation of these formulae is straightforward, but tedious, matrix algebra. The matrix inversion lemma comes in handy.