How can I convert A matrix to B matrix below?

I want to convert

A=[a11 a12 a13; 
   a22 a22 a23; 
   a31 a32 a33]

to

B=[a11 a11 a11;
   a11 a22 a22;
   a11 a22 a33]

Thank you!

Answer 1

[Jos (10584)]

B = eye(size(A)).*A
B = cumsum(B,1) + cumsum(B,2) - B

[Rengin]

Thankssss!!!

Answer 2

[Image Analyst]

Try it this way:

% This robust code works even if A is not square.
A = randi(99, 7,10) % Sample data.
B = A; % Initialize.
for k = 1 : min(size(A)) % Go only as far as we can.
  B((k+1):end, k) = A(k,k); % Send value down along column.
  B(k, (k+1):end) = A(k,k); % Send value along row to the right.
end
B % Print out to command window.

[Joseph Cheng]

Very elegant, good approach to the solution where i originally thought would be a bit overkill if it was only 3x3. Very ingenious to even go by k+1 as to not re-write the diagonals which already has the value.

[Rengin]

Thank you for your answer but i couldn't get the results which I wanted to see. In my matrix except my diagonal elements, all elements are "0". When I wrote your code, my matrix remained same.

[Image Analyst]

That would have been good to state at the start, though it doesn't really matter to the code whether the off diagonals are zero or not. The code still works. Here's proof, where I make A a diagonal matrix:

% Now, do again for A being diagonal:
numberOfRows = 7;
A = randi(99, numberOfRows, numberOfRows) % Sample data.
A = A .* eye(numberOfRows) % Zero out everything except the diagonal.
B = A; % Initialize.
for k = 1 : min(size(A)) % Go only as far as we can.
  B((k+1):end, k) = A(k,k); % Send value down along column.
  B(k, (k+1):end) = A(k,k); % Send value along row to the right.
end
B % Print out to command window.

A =
    31     0     0     0     0     0     0
     0    45     0     0     0     0     0
     0     0     8     0     0     0     0
     0     0     0     9     0     0     0
     0     0     0     0    15     0     0
     0     0     0     0     0    35     0
     0     0     0     0     0     0    42
B =
    31    31    31    31    31    31    31
    31    45    45    45    45    45    45
    31    45     8     8     8     8     8
    31    45     8     9     9     9     9
    31    45     8     9    15    15    15
    31    45     8     9    15    35    35
    31    45     8     9    15    35    42

Not sure what you meant by the matrix remaining the same. Obviously B is not the same as A and is what you wanted. And is exactly the same output as what Jos's code produces so I'm puzzled as to why you say mine didn't give the results you wanted but Jos's does. So it works fine. But whatever, if the cumsum method is easier for you, then fine.

Answer 3

[Roger Stafford]

Try this:

 B = repmat(diag(A),size(A,2));
 B = triu(B)+tril(B.',-1);

[Rengin]

Thank you for your answer but when I wrote your code, I got a such warning: Matrix dimensions must agree.

Answer 4

[Joseph Cheng]

Pretty much how you just typed it?

B = [A(1,1) A(1,1) A(1,1);...
     A(1,1) A(2,2) A(2,2);...
     A(1,1) A(2,2) A(3,3)];

or is there more to your question?

[Rengin]

I am not asking it! a11,a22,....ann my diagonal elements. Let's say I have 100x100 sized matrix and for a72-72, the rows from 72 to 100 in 72th column and the columns from 72 to 100 in 72th row have to be a72-72

Answer 5

[Andrei Bobrov]

z = diag(A);
ii = 1:numel(z);
B = z(bsxfun(@min,ii,ii'));