gernerating standard normal variable in matlab

I want to generate a vector 'z' of standard normal variable of size n. I want to create a matrix by multiplying z and z transpose, i.e. a=zz'. the matrix a should be ideally a unit matrix of size(n,n). I want to check wether the matrix is close to identity or not ?.

n = 10;  % Specify the size of the vector and matrix
z= normrnd(1,0,[n,n]);  % Generate a vector of standard normal variables


a = z * z';  % Compute the matrix by multiplying z and its transpose
threshold = 1e-6;  % Define a threshold for closeness

norm_diff = norm(a - eye(n), 'fro');  % Compute the Frobenius norm of the difference

if norm_diff < threshold
    disp('The matrix a is close to the identity matrix.');
else
    disp('The matrix a is not close to the identity matrix.');
end

The matrix a is not close to the identity matrix.

at what value of n will i get the value close to identity?

The MATLAB command says that :

r = normrnd(mu,sigma,sz1,...,szN) generates an array of normal random numbers, where sz1,...,szN indicates the size of each dimension.

help normrnd

 NORMRND Random arrays from the normal distribution.
    R = NORMRND(MU,SIGMA) returns an array of random numbers chosen from a
    normal distribution with mean MU and standard deviation SIGMA.  The size
    of R is the common size of MU and SIGMA if both are arrays.  If either
    parameter is a scalar, the size of R is the size of the other
    parameter.
 
    R = NORMRND(MU,SIGMA,M,N,...) or R = NORMRND(MU,SIGMA,[M,N,...])
    returns an M-by-N-by-... array.
 
    See also NORMCDF, NORMFIT, NORMINV, NORMLIKE, NORMPDF, NORMSTAT,
    RANDOM, RANDN.

    Documentation for normrnd
       doc normrnd

Answer 1

[Matt J]

at what value of n will i get the value close to identity?

Well, with your current code, z will always contain only ones, so the answer is none.

n = 10;  % Specify the size of the vector and matrix
z= normrnd(1,0,[n,n])

z = 10×10

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

What you probably meant to have is,

z= randn(n);
z=z-mean(z);
a=z*z'/n;

[Matt J]

But even with the corrections I made above, you have no hope of reaching the error threshold of 1e-6 when n is a realistic matrix dimension. Even if you use the max-norm instead of the Frobenius norm, and even if you had a computer that could handle n=1e7, the error in just one element of the matrix 'a' will be well above 1e-6. Here is a simple test to show that:

n=1e7;
x=randn(1,n); x=x-mean(x);
y=randn(1,n); y=y-mean(y);

varError = abs(mean(x.^2)-1) , %Error in the variance (a diagonal element a(i,i))

varError = 9.1233e-05

corrError = abs(mean(x.*y)-0) , %Error in cross-correlation (an off-diagonal a(i,j))

corrError = 3.8877e-04

[rakesh kumar]

my question is, if z_i is a standard normal varible (i,e. with mean zero and variance 1) and if I have a vector z=[z_1 z_2..........z_n] ' then theorectically zz^T should give an identity matrix. I just want to genrate that z in matlab.

[Matt J]

if I have a vector z=[z_1 z_2..........z_n] ' then theorectically zz^T should give an identity matrix.

No, that is false. If you generate millions of such vectors z and average together the z*z' matrices computed from them, then yes, you will converge to the identity matrix. However, a single such vector z will not be sufficient.

Here's another example that might make that clearer:

n=1e7;

z=normrnd(0,1,[2,n]);

a1=z(:,1)*z(:,1)'  %not close at all to eye(2)

a1 = 2×2

0.0883 0.5635 0.5635 3.5951

a2=z*z'/n          %fairly close to eye(2)

a2 = 2×2

0.9997 -0.0003 -0.0003 1.0003

[rakesh kumar]

very sorry, my theoretical assumptions was incorrect.

Answer 2

[John D'Errico]

I'm a little confused. Why would you expect the result (of z*z') would be even remotely close to an identity matrix? In fact, your claim the result should be any kind of unit matrix would be wrong.

First of all, this is NOT even true:

z= normrnd(1,0,[n,n]);  % Generate a vector of standard normal variables

z =
     1     1     1     1     1     1     1     1     1     1
     1     1     1     1     1     1     1     1     1     1
     1     1     1     1     1     1     1     1     1     1
     1     1     1     1     1     1     1     1     1     1
     1     1     1     1     1     1     1     1     1     1
     1     1     1     1     1     1     1     1     1     1
     1     1     1     1     1     1     1     1     1     1
     1     1     1     1     1     1     1     1     1     1
     1     1     1     1     1     1     1     1     1     1
     1     1     1     1     1     1     1     1     1     1
     
a =
    10    10    10    10    10    10    10    10    10    10
    10    10    10    10    10    10    10    10    10    10
    10    10    10    10    10    10    10    10    10    10
    10    10    10    10    10    10    10    10    10    10
    10    10    10    10    10    10    10    10    10    10
    10    10    10    10    10    10    10    10    10    10
    10    10    10    10    10    10    10    10    10    10
    10    10    10    10    10    10    10    10    10    10
    10    10    10    10    10    10    10    10    10    10
    10    10    10    10    10    10    10    10    10    10

that does NOT generate a vector of standard normal variates. It generates an array of size 10x10, with mean 1, and standard deviation ZERO. So when you do that, you get an array of all ONES. And therefore, the array a will be an array of all 10's.

But even if you DID generate an array of normal variates, z*z' would still NOT generate anything close to an identity matrix.

I think you are a little confused in this.

[rakesh kumar]

my question is, if z_i is a standard normal varible (i,e. with mean zero and variance 1) and if I have a vector z=[z_1 z_2..........z_n] ' then theorectically zz^T should give an identity matrix. I just want to genrate that z in matlab.

[rakesh kumar]

My question is, if z_i is a standard normal variable (i,e. with mean zero and variance 1) and if I have a vector z=[z_1 z_2..........z_n] ' then theoretically zz^T should give an identity matrix. I just want to generate that z in MATLAB.

The MATLAB command says:

r = normrnd(mu,sigma,sz1,...,szN) generates an array of normal random numbers, where sz1,...,szN indicates the size of each dimension.

help normrnd

 NORMRND Random arrays from the normal distribution.
    R = NORMRND(MU,SIGMA) returns an array of random numbers chosen from a
    normal distribution with mean MU and standard deviation SIGMA.  The size
    of R is the common size of MU and SIGMA if both are arrays.  If either
    parameter is a scalar, the size of R is the size of the other
    parameter.
 
    R = NORMRND(MU,SIGMA,M,N,...) or R = NORMRND(MU,SIGMA,[M,N,...])
    returns an M-by-N-by-... array.
 
    See also NORMCDF, NORMFIT, NORMINV, NORMLIKE, NORMPDF, NORMSTAT,
    RANDOM, RANDN.

    Documentation for normrnd
       doc normrnd

[rakesh kumar]

very sorry, my theoretical assumptions was incorrect.