Solve det(A-Ic)=O for c (there will be multiple sols) (That means A, but take away c from each number in the diagonal)

For each c solve (A-Ic)(x0,..,xn)^t=0 You will get infinitely many sols, the eigenformula will be the formula that gives all the solutions.

For example A=(1,2) (0,3) Det|1-c, 2|=(1-c)(3-c) - 0*2 |0,3-c| Is 0 when c = 1 or c = 3, those are the eigenvalues

The eigenformula for c = 1:

(1-c)x + 2y = 0 0x + (3-c)y = 0

Simplifies to y=0

Im not sure how they want to express the eigenformula, in Germany we use eigenvectors. But i assume it will be something along the lines of F1(k) = k F2(k) = 0

Or f(k) = k*|1| |0|

Then do the same for c=3