Eigen values product $ M= \left[ {
\begin{array}{ccccc}
 1 & 0 & 0 & 0 & 1\\
 0 & 1 & 1 & 1 & 0\\
 0 & 1 & 1 & 1 & 0\\
 0 & 1 & 1 & 1 & 0\\
1 & 0 & 0 & 0 & 1\\
 \end{array} } \right] $
What is the product of positive eigen values for the above matrix ?
I got the answer using normal method(by transforming to echelon form) to find the eigen values but it took me a lot of time.
Is there any shortcut or simple way to solve this ? After looking at the structure of the matrix, it seems that there is some shorter way to do it but I am not able to find it.
Can someone please help?
 A: Notably, the matrix $M$ has rank $2$, which is to say that it will have $2$ non-zero eigenvalues.  Both of these eigenvalues will be positive, since the matrix is positive semidefinite. 
Let $\lambda_i$ denote the eigenvalues of $M$. We can note that 
$$
\operatorname{tr}(M^2) = \sum_{i}\lambda_i^2\\
\operatorname{tr}(M)^2 = \left(\sum_{i}\lambda_i \right)^2 = \sum_{i, j} \lambda_i\lambda_j
$$
So that
$$
\operatorname{tr}(M)^2 - \operatorname{tr}(M^2) = \sum_{i \neq j} \lambda_i\lambda_j
$$
In other words, we can compute the product of the eigenvalues of this rank-2 positive definite matrix as
$$
\lambda_1 \lambda_2 = \frac 12\left[\operatorname{tr}(M)^2 - \operatorname{tr}(M^2) \right]
$$

On the other hand, we can find a permutation matrix $P$ such that
$$
PMP^T = \pmatrix{1&1\\1&1\\&&1&1&1\\&&1&1&1\\&&1&1&1}
$$
Since this matrix is block-diagonal, it isn't difficult to see that the non-zero eigenvalues of the individual blocks (and hence of the entire matrix) will be $2$ and $3$ (respectively).
A: 
Obviously $M$ has rank two. Therefore M has only two non-negative eigenvalues(count with multplicity).  

It is easy to check that $v_2 =(1,0,0,0,1)^T$ is an eigenvector associate to the eigenvalue 2. and $v_3 =(0,1,1,1,0)^T$ is an eigenvector associate to the eigenvalue 3. 
 $$Mv_2 =2v_2~~~and ~~~~Mv_3 =3v_3$$

Answer is $2\times 3 = 6.$


