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I am trying to solve a mathematical problem related to matrices using numpy as shown below:

I am really finding it hard to represent this kind matrix structure using numpy. I really donot want to type these values because I want to understand how this kind of structures are represented using python. Consider the empty places as zero. Thank you for great help.

I am trying to solve a mathematical problem related to matrices using numpy as shown below:

I am really finding it hard to represent this kind matrix structure using numpy. I really donot want to type these values because I want to understand how this kind of structures are represented using python. Consider the empty places as zero. Thank you for great help.

• python
• numpy
• matrix

Share Improve this question Follow edited Jan 31 at 15:02 ThomasIsCoding 103k99 gold badges3636 silver badges101101 bronze badges asked Jan 30 at 20:18 mchaudh4mchaudh4 21711 silver badge99 bronze badges 2

• To clarify, this is not tridiagonal (tridiagonal only has nonzero values on the main diagonal and the two off-diagonals). Can you clarify what you mean by "understand[ing] how this kind of structures are represented using python"? – jared Commented Jan 30 at 20:47
• Thank you for the response @jared. There are ways to put a complete diagonal or off-diagonal to same value, but here there are three continuous -1, and then there is zero. Not sure what this kind of structure should be called. – mchaudh4 Commented Jan 30 at 20:59

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2 Answers 2

Sorted by: Reset to default 2

The matrix has to be decomposed into several parts. First, the middle region forms a block diagonal matrix, where each block is a 4x4 Toeplitz matrix.

# makes a shifted diagonal matrix
def E(n, v, k):
    return v * np.eye(n, k=k)

def toeplitz_block(n, v_list, k_list):
    return sum(E(n, v, k) for v, k in zip(v_list, k_list))

Then we can do the following:

n = 4 # size of block
m = 3 # how many blocks

# Make block diagonal matrix A
A = np.zeros((m, n, m, n))
u, v = np.diag_indices(m)
A[u, :, v, :] = toeplitz_block(n, [-1, 3, -1], [-1, 0, 1])
A = A.reshape(m * n, m * n)

print(A.astype(int))

# Output:
[[ 3 -1  0  0  0  0  0  0  0  0  0  0]
 [-1  3 -1  0  0  0  0  0  0  0  0  0]
 [ 0 -1  3 -1  0  0  0  0  0  0  0  0]
 [ 0  0 -1  3  0  0  0  0  0  0  0  0]
 [ 0  0  0  0  3 -1  0  0  0  0  0  0]
 [ 0  0  0  0 -1  3 -1  0  0  0  0  0]
 [ 0  0  0  0  0 -1  3 -1  0  0  0  0]
 [ 0  0  0  0  0  0 -1  3  0  0  0  0]
 [ 0  0  0  0  0  0  0  0  3 -1  0  0]
 [ 0  0  0  0  0  0  0  0 -1  3 -1  0]
 [ 0  0  0  0  0  0  0  0  0 -1  3 -1]
 [ 0  0  0  0  0  0  0  0  0  0 -1  3]]

To get the final desired matrix, we can just add another Toeplitz matrix that has the diagonal of -1s.

B = A + E(n * m, -1, -4)
print(B.astype(int))

# Output:
[[ 3 -1  0  0  0  0  0  0  0  0  0  0]
 [-1  3 -1  0  0  0  0  0  0  0  0  0]
 [ 0 -1  3 -1  0  0  0  0  0  0  0  0]
 [ 0  0 -1  3  0  0  0  0  0  0  0  0]
 [-1  0  0  0  3 -1  0  0  0  0  0  0]
 [ 0 -1  0  0 -1  3 -1  0  0  0  0  0]
 [ 0  0 -1  0  0 -1  3 -1  0  0  0  0]
 [ 0  0  0 -1  0  0 -1  3  0  0  0  0]
 [ 0  0  0  0 -1  0  0  0  3 -1  0  0]
 [ 0  0  0  0  0 -1  0  0 -1  3 -1  0]
 [ 0  0  0  0  0  0 -1  0  0 -1  3 -1]
 [ 0  0  0  0  0  0  0 -1  0  0 -1  3]]

Here is another option

n = 12
msk = (v:=np.arange(n))-v[:,None]
A = 3*(msk==0) - (np.logical_or(abs(msk)==1,msk==-4))

such that

A=
 [[ 3 -1  0  0  0  0  0  0  0  0  0  0]
 [-1  3 -1  0  0  0  0  0  0  0  0  0]
 [ 0 -1  3 -1  0  0  0  0  0  0  0  0]
 [ 0  0 -1  3 -1  0  0  0  0  0  0  0]
 [-1  0  0 -1  3 -1  0  0  0  0  0  0]
 [ 0 -1  0  0 -1  3 -1  0  0  0  0  0]
 [ 0  0 -1  0  0 -1  3 -1  0  0  0  0]
 [ 0  0  0 -1  0  0 -1  3 -1  0  0  0]
 [ 0  0  0  0 -1  0  0 -1  3 -1  0  0]
 [ 0  0  0  0  0 -1  0  0 -1  3 -1  0]
 [ 0  0  0  0  0  0 -1  0  0 -1  3 -1]
 [ 0  0  0  0  0  0  0 -1  0  0 -1  3]]

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