Simple way to find determinant of a given matrix Consider the $(m-1)\times (m-1)$ matrix $M$ shown below.
  $$ M=
  \left[ {\begin{array}{cc}
   1 & 1/2 & 1/2 & \ldots & 1/2\\
   1/2 & 1 & 1/2 & \ldots & 1/2\\
\vdots \\
1/2 & 1/2 & 1/2 & \dots & 1
  \end{array} } \right]$$
I would like to show that $\det (M)=\frac{m}{2^{m-1}}$. I have tried induction, but the computation gets rather messy. When $m=2$, $\det(M)=1=2/2$, and when $m=2$, $\det(M)=1-1/2^2=3/2^2$. 
I know some people are very good at seeing tricks to compute the determinant of a matrix like this without induction. Any ideas?
Thanks!
 A: The determinant is the product of the eigenvalues. We know that $\frac12$ is an eigenvalue with multiplicity at least $m-2$, because $M - \frac12I$ has rank $1$. In addition, the sum of the eigenvalues is the trace of $M$, which is $m-1$. So if $x$ is the remaining eigenvalue, then $x + \frac12(m-2) = m-1$, which means that $x = \frac m2$.
Taking the product of the eigenvalues, we get $\frac m2 \cdot \left(\frac12\right)^{m-2} = \frac{m}{2^{m-1}}$.

Also, Sylvester's determinant identity applies to $2$ times this matrix. If $\mathbf u$ is the vector of all $1$'s, then $2M = I + \mathbf{u}\mathbf{u}^T$, so $\det(2M) = \det(I + \mathbf{u}\mathbf{u}^T) = 1 + \mathbf{u}^T\mathbf{u} = m$. Therefore $\det(M) = \frac1{2^{m-1}} \det(2M) = \frac{m}{2^{m-1}}$.
A: More generally, consider the $n\times n$ matrix
$$
M(a)=\begin{bmatrix}
a & a & a & \dots & a \\
a & a & a & \dots & a \\
\vdots & \vdots & \ddots & \dots & \vdots \\
a & a & a & \dots & a
\end{bmatrix}
$$
The characteristic polynomial of $M(a)$ is
$$
\det(M(a)-XI)=(0-X)^{n-1}(na-X)
$$
Can you see why?
What value can you give to $X$ in order to have your given matrix?
