I have most of the times come across authors who given a characteristic matrix, proceed to find its eigenvalues by solving for the $\left | A - \lambda I \right |=0 $ and then solving the characteristic polynomial.
While playing around with the concept, I reasoned, that $A-\lambda I$ will always have a rank less than 'n' where $n$ is the number of variables for $\mathbf{X}$ to have non trivial solution.
Thus I went about solving for the echelon form of the coefficient matrix $A-\lambda I$ in the way, proving that in the echelon form, one of the rows must vanish to satisfy the rank condition, and thus equating the last row, last column to 0, which almost always gave me the characteristic polynomial.
My question is:
has anyone tried solving it this way?
does this way offer any computational advantage over calculation of determinants when 'n' increases?
Any thoughts regarding the same.