How to stop unwanted rows being added when constructing a matrix in a for loop

Hi there,

just a simple question I think.

I am trying to construct a matrix in a quick way using a for loop. Because the values for diagonals are the same I can using indexing like (i,i) and (i+1,i) etc. However, apart from the first diagonal I input (i,i), the second diagonal (i,i+1) adds an extra row and column to my original matrix.

Here is my code:

clear, clc, close all

mat = zeros(7,7)
n = size(mat,1)

for i = 1:n
    mat(i,i) = 6
    mat(i,i+1) = -4
end

If you run the code, the diagonal containing -4 adds on a extra row and column at the end; making a 7x7 matrix to an 8x8. This is very frustraing and I do not want this!

Can I ask what its the trick to getting around this, please?

Many thanks

Scott

[Stephen23]

n = 7;
toeplitz([6,zeros(1,n-1)],[6,-4,zeros(1,n-2)])

ans = 7x7

6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6

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Answer 1

[Star Strider]

The reason is that there are the same number of ‘6’ and ‘-4’ being created in the loop. There needs to be one less ‘-4’.

clear, clc, close all

mat = zeros(7,7);
n = size(mat,1);

for i = 1:n
    mat(i,i) = 6;
    mat(i,i+1) = -4;
end



mat

mat = 7x8

6 -4 0 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 0 6 -4

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N = 7;
mat2 = diag(ones(1,N)*6);                               % Use 'diag'
mat2 = + mat2 + diag(ones(1,N-1)*-4, 1)

mat2 = 7x7

6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6

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One way to get around that iis to use the diag function to creeate the matrix.

.

Answer 2

[ScottB]

mat = zeros(7,7)
n = size(mat,1)

for i = 1:n
    mat(i,i) = 6
    if i ==7
      else
      mat(i,i+1) = -4
    end
end

mat =

6 -4 0 0 0 0 0

0 6 -4 0 0 0 0

0 0 6 -4 0 0 0

0 0 0 6 -4 0 0

0 0 0 0 6 -4 0

0 0 0 0 0 6 -4

0 0 0 0 0 0 6

[Voss]

Or

mat = zeros(7,7);
n = size(mat,1);
for i = 1:n
    mat(i,i) = 6;
    if i ~= n
        mat(i,i+1) = -4;
    end
end

mat

mat = 7x7

6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6

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Answer 3

[Steven Lord]

Since you're creating matrices with diagonal bands, consider using the diag or spdiags functions.

n = 7;
mainDiagonal = diag(6*ones(n, 1))

mainDiagonal = 7x7

6 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 6

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upperDiagonal = diag(-4*ones(n-1, 1), 1)

upperDiagonal = 7x7

0 -4 0 0 0 0 0 0 0 -4 0 0 0 0 0 0 0 -4 0 0 0 0 0 0 0 -4 0 0 0 0 0 0 0 -4 0 0 0 0 0 0 0 -4 0 0 0 0 0 0 0

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A = mainDiagonal + upperDiagonal

A = 7x7

6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6

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Note that when I created upperDiagonal I used a vector of all -4 values that had length one shorter than the vector of 6 values I used to create mainDiagonal. This made it so upperDiagonal was n-by-n rather than (n+1)-by-(n+1) as it would have been had I used -4*ones(n, 1).

B = diag(-4*ones(n, 1), 1)

B = 8x8

0 -4 0 0 0 0 0 0 0 0 -4 0 0 0 0 0 0 0 0 -4 0 0 0 0 0 0 0 0 -4 0 0 0 0 0 0 0 0 -4 0 0 0 0 0 0 0 0 -4 0 0 0 0 0 0 0 0 -4 0 0 0 0 0 0 0 0

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For bands below the diagonal, use a negative value as the second input.

lowerDiagonal = diag(-8*ones(n-1, 1), -1)

lowerDiagonal = 7x7

0 0 0 0 0 0 0 -8 0 0 0 0 0 0 0 -8 0 0 0 0 0 0 0 -8 0 0 0 0 0 0 0 -8 0 0 0 0 0 0 0 -8 0 0 0 0 0 0 0 -8 0

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Here are the numbers for which band contains each element of the matrix.

bandNumbers = toeplitz(0:-1:-(n-1), 0:(n-1))

bandNumbers = 7x7

0 1 2 3 4 5 6 -1 0 1 2 3 4 5 -2 -1 0 1 2 3 4 -3 -2 -1 0 1 2 3 -4 -3 -2 -1 0 1 2 -5 -4 -3 -2 -1 0 1 -6 -5 -4 -3 -2 -1 0

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For spdiags you don't have to create each band individually, though the syntax is a little bit more complicated. The command below puts 6 on the 0 (main) diagonal of S and -4 on the +1 (upper) diagonal, and makes S an n-by-n matrix.

 S = spdiags([6 -4], [0 1], n, n)

S = 
   (1,1)        6
   (1,2)       -4
   (2,2)        6
   (2,3)       -4
   (3,3)        6
   (3,4)       -4
   (4,4)        6
   (4,5)       -4
   (5,5)        6
   (5,6)       -4
   (6,6)        6
   (6,7)       -4
   (7,7)        6

Since S is sparse it's displayed slightly differently. Convert it to full and it looks like A above.

F = full(S)

F = 7x7

6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6 -4 0 0 0 0 0 0 6

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Answer 4

[Scott Banks]

Briiliant, thanks guys for all youre help!