用numpy表示三对角矩阵

矩阵必须分解成几个部分。首先,中间区域形成块对角矩阵,其中每个块是4x4 Toeplitz矩阵。

# 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))

然后我们可以执行以下操作:

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]]

为了得到最终的期望矩阵,我们可以添加另一个对角线为-1的Toeplitz矩阵。

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]]

这是另一种选择

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

这样

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]]