Determinant of symmetric matrix with non negative integer element

[golekjwb]

Let \begin{equation*}
A=%
\begin{bmatrix}
0 & 1 & \cdots & n-1 & n \\
1 & 0 & \cdots & n-2 & n-1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
n-1 & n-2 & \cdots & 0 & 1 \\
n& n-2 & \cdots & 1 & 0%
\end{bmatrix}%
\end{equation*}.
How can you prove that det(A)=[(-1)^n][n][2^(n-1)]? Thanks.

 

[golekjwb]

Try n=4

 

[chiro]

Try n=4

Hey golekjwb and welcome to the forums.

For the general proof I would use an induction argument. The differences between say n = k and n = k + 1 has to do with evaluating one extra minor determinant for that extra row and you would show that under a simplification that the formula is correct.

For a specific n=4, just evaluate the determinant for that particular dimension for your particular matrix, expand out and see what you get.

 

[tiny-tim]

welcome to pf!

hi golekjwb! welcome to pf!

have you tried adding or subtracting rows or columns, to get a simpler matrix?

 

[blue_raver22]

I would approach this problem by first understanding the properties of symmetric matrices and their determinants. A symmetric matrix is a square matrix that is equal to its transpose, meaning that the elements above and below the main diagonal are symmetrical.

To prove the determinant of a symmetric matrix with non-negative integer elements, we can use the cofactor expansion method. The determinant of a matrix is equal to the sum of the products of the elements in any row or column multiplied by their corresponding cofactors.

In this case, we can choose the first column to expand the determinant. The first element in the first row is 0, so the corresponding cofactor is (-1)^{1+1} = 1. The determinant of the remaining (n-1)x(n-1) matrix is denoted as det(A').

Therefore, the determinant of the first column is 0 x det(A') = 0. We can continue this process for the remaining columns, where the determinant of the second column is 1 x det(A''), the third column is 2 x det(A'''), and so on.

Since the elements in each column are multiplied by a different number, we can factor out the common term n! from the determinant. This gives us det(A) = n! x det(A').

Next, we can use the fact that det(A') is equal to the determinant of a (n-1)x(n-1) symmetric matrix with non-negative integer elements. We can repeat the process above, choosing the first column to expand the determinant. This time, the first element in the first row is 1, so the corresponding cofactor is (-1)^{1+1} = 1. The determinant of the remaining (n-2)x(n-2) matrix is denoted as det(A'').

Continuing this process, we can see that det(A') = (n-1) x (n-2) x det(A''). We can factor out another common term (n-1)! from the determinant, giving us det(A) = n! x (n-1)! x det(A'').

We can continue this process until we reach a 2x2 matrix, where the determinant is simply the product of the elements on the main diagonal (0 x 0 - 1 x (n-1) = - (n-1)). Therefore, det(A) = n