Calculating the Determinant of a Circulant Matrix

The given matrix

Matrix M

is of the form

[mathbf{M} x - y mathbf{I} - y mathbf{j} mathbf{j}^T]

where (mathbf{I}) is the identity matrix and (mathbf{j}) is the vector of all ones. The determinant of (mathbf{M}) can be found using the matrix determinant lemma. This states that for any matrix (mathbf{A}), vector (mathbf{u}), and vector (mathbf{v}), the determinant is given by:

[det(mathbf{A} mathbf{u} mathbf{v}^T) det(mathbf{A}) (1 mathbf{v}^T mathbf{A}^{-1} mathbf{u})]

In our case, we have (mathbf{A} x - y mathbf{I}), (mathbf{u} y mathbf{j}), and (mathbf{v} mathbf{j}). Let's follow the steps to compute the determinant:

Apply the matrix determinant lemma: Determine (det(x - y mathbf{I})) first. Since (mathbf{I}) is the identity matrix, the determinant is simply: Use the lemma to find the determinant of (mathbf{M}) by calculating (1 mathbf{j}^T (x - y mathbf{I})^{-1} mathbf{j}):

We know that:

[mathbf{j}^T mathbf{j} y^n]

Also, the inverse of (x - y mathbf{I}) is:

[(x - y mathbf{I})^{-1} frac{1}{x - y} mathbf{I}]

Therefore, the term above simplifies to:

[mathbf{j}^T (x - y mathbf{I})^{-1} mathbf{j} frac{1}{x - y} y^n frac{y^n}{x - y}]

Thus, the determinant is:

[det(mathbf{M}) (x - y) left(1 frac{y^n}{x - y}right) (x - y) left(frac{x - y y^n}{x - y}right) x - y y^n - y (x - y)^{n-1} x y^{n-1}]

Circulant Matrices

The matrix in question is a circulant matrix. Circulant matrices are square matrices that begin with an arbitrary first non-zero row and then subsequent rows are obtained by shifting the previous row to the right one entry and cycling the last entry around to the front. A generic (4 times 4) example is:

[text{circ}(a, b, b, b) begin{bmatrix} a b b b b a b b b b a b b b b a end{bmatrix}]

Circulants have a number of interesting properties. One is that they are all diagonalizable, meaning for each positive integer (n), there is a single basis of eigenvectors that diagonalizes all (n times n) circulants. A second property is that the eigenvalues of circulants can be found easily. For each circulant (C), there is a polynomial (r_C(x)) (called the representing polynomial) that can be directly read off from the circulant. Substituting the (n)th roots of unity into (r_C(x)) gives the eigenvalues of (C). The determinant of (C) is given by the product of these eigenvalues:

[det(C) r_C(1) r_C(omega) r_C(omega^2) dots r_C(omega^{n-1})]

To illustrate, consider the circulant (C text{circ}(a, b, b, dots, b)). The representing polynomial (r_C(x)) is:

[r_C(x) a b x b x^2 dots b x^{n-1} a b x (x^{n-1} x^{n-2} dots x) a frac{b (x^n - x)}{x - 1}]

When (x eq 1), the representing polynomial simplifies to:

[r_C(x) a b frac{x^n - x}{x - 1}]

When (x 1), the term simplifies directly to:

[r_C(1) a (n-1)b]

Since the eigenvalues are (r_C(omega)) as (omega) ranges over all (n)th roots of unity, the determinant is the product of these eigenvalues:

[det C left(a (n-1)bright) (a - b)^{n-1}]

For further reading and in-depth knowledge about circulants, refer to existing literature and resources on the topic.