Small Matrix Eigenvalue

There is a shortcut for any square matrix. The trace of a matrix is equal to the sum of its eigenvalues. See my note Product and Sum of Eigenvalues .

The Eigenvalues satisfy the condition $det(A-\lambda { I }_{ n })=0$ where $A$ is a matrix, $\lambda$ is an Eigenvalue, and ${I}_{n}$ is the $n \times n$ identity matrix

Thus, $\begin{vmatrix} 4-\lambda & 6 \\ 3 & 8-\lambda \end{vmatrix}=0$

${ \lambda }^{ 2 }-12\lambda +32-18\quad =\quad 0$

The sum of the Eigenvalues is the sum of the roots, or $12$ .

Notice that the sum of the Eigenvalues is the sum of the trace elements (the elements in the downward sloping diagonal)

just get the trace of its diagonal which is 4 and 8

4 + 8 = 12

1. just add 4 and 8 because the charasteristic polynomial will (x-4)(x-8)-18=0. but we just need the sum of the eigenvalues, and the sum of the roots can be based on the equation alone. x^2-12x+14=0. the sum of the roots is 12.