Using while loops in matrices

I am trying to use while loop to change the diagonal entries of a square matrix rand(10) to 1, and other entries to zero

This code below is changing the whole entries to 1, i am stucked.

m= 1:10
n= 1:10
A = rand(10)
B = size (A)
while m==n    
    A(m,n) = 1;
    if not (m==n)
        A(m,n) = 0;
    end 
    break
end 
A

[Torsten]

A = rand(10);

A = eye(10);

Answer 1

[Pin-Hao Cheng]

Let me know if this is what you are looking for. Happy to answer any further questions!

A = rand(10)

A = 10×10

0.1406 0.9234 0.5213 0.0271 0.2317 0.3967 0.3776 0.4847 0.3080 0.0222 0.4363 0.4473 0.2888 0.3199 0.6071 0.3528 0.3311 0.5584 0.4040 0.2190 0.2656 0.6738 0.6987 0.3731 0.8456 0.9905 0.1035 0.4882 0.4701 0.2163 0.0655 0.9590 0.1978 0.6575 0.6074 0.2159 0.2553 0.3445 0.9210 0.6419 0.8856 0.9335 0.8339 0.5396 0.2503 0.0700 0.4051 0.7461 0.3052 0.1634 0.4934 0.4743 0.6264 0.2365 0.6042 0.4961 0.4183 0.7427 0.0465 0.3843 0.2964 0.5497 0.6846 0.4194 0.2852 0.9456 0.7047 0.9608 0.4107 0.6736 0.7785 0.3134 0.1016 0.5672 0.2764 0.7794 0.3809 0.2773 0.2288 0.1847 0.6817 0.5795 0.6303 0.4710 0.6249 0.1779 0.3139 0.5706 0.5860 0.4254 0.6862 0.1720 0.4387 0.9392 0.0916 0.3381 0.4475 0.3649 0.5224 0.7542

for m = 1:10        % loop through rows
    for n = 1:10    % loop through columns
        if m == n   % check if it's diagonal el
            A(m,n) = 1;
        else
            A(m,n) = 0;
        end

    end
end

A

A = 10×10

1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1

[Olabayo]

This is using the FOR loop. I want to use the WHILE loop for achive this same result

[Askic V]

How about this:

% Initialize matrix A with random values
N = 5;
A = rand(N)

A = 5×5

10 2 6 7 8 6 2 7 9 7 3 6 6 2 4 2 6 6 4 9 1 7 3 1 1

% processing the elements
while N > 0
    A(N,:) = 0;
    A(N,N) = 1;
    N = N-1;
end
A

A = 5×5

1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1

[Walter Roberson]

A = rand(10)

A = 10×10

0.5267 0.0485 0.6881 0.0751 0.5088 0.9623 0.4144 0.0369 0.5932 0.1203 0.0743 0.4432 0.1701 0.8212 0.2084 0.4752 0.3201 0.4127 0.1457 0.4379 0.4294 0.8825 0.5937 0.8722 0.1412 0.2457 0.6319 0.1385 0.0125 0.1849 0.1902 0.2588 0.2948 0.1180 0.3554 0.0058 0.1091 0.1462 0.6565 0.7121 0.3901 0.4582 0.6801 0.5591 0.1280 0.3830 0.4597 0.2188 0.6096 0.7307 0.7630 0.4357 0.6976 0.8166 0.7730 0.5614 0.6620 0.1153 0.8800 0.3424 0.9485 0.5221 0.1301 0.3194 0.6058 0.7353 0.1995 0.8348 0.8918 0.8874 0.3259 0.7321 0.2832 0.4322 0.1251 0.5902 0.5059 0.1353 0.5748 0.5123 0.2665 0.2283 0.9713 0.8249 0.1909 0.8411 0.5445 0.3293 0.1103 0.8057 0.1126 0.5050 0.2498 0.4914 0.5326 0.4910 0.3327 0.8345 0.2003 0.1004

m = 1;
while m <= 10        % loop through rows
    n = 1;
    while n <= 10    % loop through columns
        if m == n   % check if it's diagonal el
            A(m,n) = 1;
        else
            A(m,n) = 0;
        end
        n = n + 1;
    end
    m = m + 1;
end

A

A = 10×10

1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1

[Pin-Hao Cheng]

Was gonna comment exactly what @Walter Roberson showed. That should solve your problem :)

[Olabayo]

Thank you for this.. this solves it

[Jan]

The pattern:

if m == n
    A(m,n) = 1;
else
    A(m,n) = 0;
end

can be abbreviated in general to:

A(m, n) = (m == n);

[Walter Roberson]

The whole thing abbreviates to a call to eye() and size()