Prove determinant of $n \times n$ matrix is $(a+(n-1)b)(a-b)^{n-1}$?

Question

Prove $\det(A)$ is $(a+(n-1)b)(a-b)^{n-1}$ where $A$ is $n \times n$ matrix with $a$'s on diagonal and all other elements $b$, off diagonal.

Answer 1

Note that for $\lambda=a-b$ we have $$ A-\lambda I = \begin{bmatrix} b & \cdots & b \\ \vdots & \ddots & \vdots \\ b & \cdots & b \end{bmatrix}\tag{1} $$ The matrix in $(1)$ has rank $1$ so its nullspace has dimension $n-1$. Hence $\lambda_1=a-b$ is an eigenvalue of $A$ whose geometric multiplicity is $n-1$.

Now, note that for $\lambda=a+(n-1)b$ we have $$ A-\lambda I= \begin{bmatrix} (1-n)b & \cdots & b \\ \vdots & \ddots & \vdots \\ b & \cdots & (1-n)b \end{bmatrix}\tag{2} $$ The columns of the matrix in $(2)$ sum to $0$ so it has a nontrivial nullspace. That is, $\lambda_2=a-(n-1)b$ is an eigenvalue for $A$. Because $\lambda_1$ has geometric multiplicity $n-1$, the geometric multiplicity of $\lambda_2$ must be $1$. Hence the characteristic polynomial of $A$ is $$ \det(A-\lambda I)=(\lambda_1-\lambda)^{n-1}(\lambda_2-\lambda)\tag{3} $$ Finally, plugging in $\lambda=0$ into $(3)$ gives the result.

Answer 2

For any square matrix with one value on the diagonal and another value everywhere else, a consistent pattern of (orthogonal but not orthonormal) eigenvectors for the $n$ by $n$ case can be read from the columns of $$ \left( \begin{array}{rrrrrrrrrr} 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9 \end{array} \right). $$ Note: I made this up. My own self.

Answer 3

Let $A=\begin{bmatrix} a & b & b & b & \ldots & b \\ b & a & b & b & \ldots & b \\ b & b & a & b & \ldots & b \\ b & b & b & a & \ldots & b \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ b & b & b & b & \ldots & a \end{bmatrix}$ a $n\times n$ matrix, cause adding/substracting a scalar multiple of a row to another row doesn't change the value of the determinant of a matrix we can add rows 2nd, 3rd, 4th, $\dots$ and $n$th to the first row, so $$\det (A) = \det \begin{bmatrix} a+(n-1)b & a+(n-1)b & a+(n-1)b & a+(n-1)b & \ldots & a+(n-1)b \\ b & a & b & b & \ldots & b \\ b & b & a & b & \ldots & b \\ b & b & b & a & \ldots & b \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ b & b & b & b & \ldots & a \end{bmatrix}$$ Now, we can factor out $a+(n-1)b$ from the first row $$\det (A) = \left[a+(n-1)b\right]\det \begin{bmatrix} 1 & 1 & 1 & 1 & \ldots & 1 \\ b & a & b & b & \ldots & b \\ b & b & a & b & \ldots & b \\ b & b & b & a & \ldots & b \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ b & b & b & b & \ldots & a \end{bmatrix}$$ Also, if we substract $b$ times row 1 to rows 2nd, 3rd, 4th, $\ldots$ and $n$th we get $$\det (A) = \left[a+(n-1)b\right]\det \begin{bmatrix} 1 & 1 & 1 & 1 & \ldots & 1 \\ 0 & a-b & 0 & 0 & \ldots & 0 \\ 0 & 0 & a-b & 0 & \ldots & 0 \\ 0 & 0 & 0 & a-b & \ldots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \ldots & a-b \end{bmatrix}$$ The last matrix is a lower triangular matrix and its determinant is the product of elements in the main diagonal. Therefore $$\det (A) = \left[a+(n-1)b\right](a-b)^{n-1}$$