How do I populate a 100 x 100 [A] matrix given rules seen below? Then use spy(A) to show non-zero entries?

      1, if i=j
Aij = 2, if i=j+1 or i=j-1
      0, otherwise

Answer 1

[Steven Lord]

If you want this to be a sparse matrix, see the second example on the documentation page for the spdiags function. It's not exactly what you want, but you should be able to adapt it.

Answer 2

[Stephen23]

Method one: toeplitz (simple to resize):

>> V = zeros(1,10);
>> V(1) = 1;
>> V(2) = 2;
>> M = toeplitz(V,V)
M = 
 1   2   0   0   0   0   0   0   0   0
 2   1   2   0   0   0   0   0   0   0
 0   2   1   2   0   0   0   0   0   0
 0   0   2   1   2   0   0   0   0   0
 0   0   0   2   1   2   0   0   0   0
 0   0   0   0   2   1   2   0   0   0
 0   0   0   0   0   2   1   2   0   0
 0   0   0   0   0   0   2   1   2   0
 0   0   0   0   0   0   0   2   1   2
 0   0   0   0   0   0   0   0   2   1

Method two: eye and indexing:

>> M = eye(10);
>> M(2:11:end) = 2;
>> M(11:11:end) = 2
M =
 1   2   0   0   0   0   0   0   0   0
 2   1   2   0   0   0   0   0   0   0
 0   2   1   2   0   0   0   0   0   0
 0   0   2   1   2   0   0   0   0   0
 0   0   0   2   1   2   0   0   0   0
 0   0   0   0   2   1   2   0   0   0
 0   0   0   0   0   2   1   2   0   0
 0   0   0   0   0   0   2   1   2   0
 0   0   0   0   0   0   0   2   1   2
 0   0   0   0   0   0   0   0   2   1

[Stephan]

Always elegant solutions. I always learn from your answers. I propose to do the indexing for case 2 dynamically on the size of M:

M = eye(10);
M(2:length(M)+1:end) = 2;
M(length(M)+1:length(M)+1:end) = 2;

[Stephen23]

@Stephan Jung: I always advise to avoid length, as what dimension it measures changes depending on the size of the input array. Ugh! There is nothing that length can do that size and numel can't do better, so they are worth getting into the habit of using (even if length works in this case).

For dynamically sizing the indexing, I would just define the value beforehand:

N = 10;
M = eye(N);
M(2:N+1:end) = 2;
M(N+1:N+1:end) = 2;

Avoiding lines of code does not always lead to a neater or clearer solution.

[Stephan]

Thanks for the feedback, I will consider this in the future.

[John Steichen]

@stephen Cobeldick can you explain what is happening in the lines of code where it changes the value? Its different than when you would go and change a single value in a matrix by specifying the row and column.

[Stephen23]

"...can you explain what is happening in the lines of code where it changes the value?"

I used basic linear indexing. Linear indexing is one of the three kind of indexing that MATLAB supports. It works columnwise and uses a single index to access all elements of any array. YOu can learn about linear indexing by reading the MATLAB documentation:

https://www.mathworks.com/help/matlab/math/array-indexing.html

https://blogs.mathworks.com/steve/2008/02/08/linear-indexing/

https://www.mathworks.com/company/newsletters/articles/matrix-indexing-in-matlab.html

Because each column contains 10 elements (in my examples), the step size for the linear index has to be 11, so as to move down one row in each subsequent column.