Determinant of a N symmetric square matrix with diagonal 1

Question

What is the determinant of a symmetric $n \times n$ matrix with all diagonals be 1 and all others are $\rho$ (yes correlation matrix)?

Anyone can tell me a method to work it out elegantly?

Thanks!

Answer 1

Let $A$ be an $n\times n$ matrix of your desired form. Let us form the matrix of all ones, call it $J$. Then we can represent $A$ as $$A=\rho J - (\rho - 1)I$$ The determinant of the above matrix is $$\det(A) = \rho^n\det\left(J - \frac{\rho - 1}{\rho}I\right)$$ Letting $\lambda = \frac{\rho - 1}{\rho}$, the latter determinant is precisely the characteristic polynomial of $J$, easily seen as $$p(\lambda)=(-1)^{n}\lambda^{n-1}(\lambda - n)$$ Some simplifying then gives the determinant as $$\det(A)=\left(1-\rho \right)^{n-1}\left(1 + \rho n-\rho\right)$$

Answer 2

Let $M_n$ be the matrix with matrix elements $$M_{ij}= \cases{1 & $i=j$\cr \rho & $i\not=j$}$$ Also let $N_n$ be the $n \times n$ matrix, obtained by $M_n$ by replacing first element of the first row with $\rho$.

Applying Laplace's method to the first row of $M_n$ and $N_n$: $$\begin{eqnarray} \det M_n &=& \det M_{n-1} - (n-1) \rho \cdot \det N_{n-1} \\ \det N_n &=& \rho \cdot \det M_{n-1} - (n-1) \rho \cdot \det N_{n-1} \end{eqnarray} $$ with $\det M_2 = 1 - \rho^2$ and $\det N_2 = \rho(1-\rho)$. This gives $$ \det(M_n) = (1-\rho)^{n-1}(1+(n-1) \rho)\qquad \det N_n = \rho (1-\rho)^{n-1} $$

Answer 3

The matrix you defined is $A = \rho J - (\rho - 1)I.$ Since $J$ has spectrum $\{0^{(n-1)}, n^{(1)}\},$ $A$ has spectrum $\{(1-\rho)^{(n-1)}, (\rho(n-1) + 1)^{(1)}\}.$

The determinant is the product of the eigenvalues, so $\det A = (1-\rho)^{n-1}(\rho(n-1)+1).$

Answer 4

One trick of applying Laplace's method to $M_n$ and $N_n$ to derive the recursive formulations is to perform the operation of switching rows (and then changing signs of the corresponding determinants accordingly).

Finding $det(M_n)$ and $det(N_n)$, on the other hand, can be done by first observing their patterns, and then using induction. Here let us take $det(M_n)$ as an example, skip the first part of the induction (cases for $n=1$ and $n=2$) and focus on the second part:

Suppose for $i=k$, the following two equations hold:

$det(M_i)=(1-\rho)^{i-1}(1+(i-1)\rho)$

$det(N_i)=\rho(1-\rho)^{i-1}$

Then for $i=k+1$, we have:

$RHS=det(M_k)-k\rho\bullet det(N_k) = (1-\rho)^{k-1}(-k\rho^2+(k-1)\rho+1)$;

Therefore,

$RHS=(1-\rho)^{k-1}(k\rho+1)(1-\rho)=(1-\rho)^{k}(1+k\rho)=LHS=det(M_{k+1})$