1) Find eigenvalues for the matrix A=((3,-2),(2,-2)). Here and later first inner parentheses denote first row, second--second row. In order to do it, one need to solve secular equation det|A-k*I|=0, where k is the eigenvalue to be found and I is unit matrix,
I=((1,0),(0,1)). In our case we will obtain:
det|((3-k,-2),(2,-2-k))|=0. From this we will get:
(3-k)(-2-k)-2*(-2)=0 or (k-3)*(k+2)+4=0 or k2-k-2=0
Solutions of this quadratic equation are given by:
2) Now, let us find eigenvectors corresponding to each of the eigenvalues. We plug in first eigenvalue, k=2, into the matrix equation above to obtain:
K1=A-k1*I=A-2I=((1,-2),(2,-4)); Now we need to solve K1(a1,a2)=0 for vector (a1,a2). If we use the explicit form of matrix K1, we will obtain two identical equations:
a1-2a2=0 and 2a1-4a2=0. Those have infinitely many solutions, corresponding to infinitely many vectors, which all point in the same direction, but differ in length. Let us pick the most obvious solution, a1=2;
a2=1; Thus, first eigenvector is (2,1). Often eigenvectors are normalized, then this eigenvector would be (2,1)/√5.
Next, we do the same for the other eigenvalue, k2. Upon substitution and solving the equation, we will obtain the second eigenvector, (1,2).
3) General solution to a differential equation in question is in the form:
C1*(first eigenvector)*Exp[k1t]+C2*(second eigenvector)*Exp[k2t]
Here k1 shall correspond to its eigenvector and so shall k2! Do not mix them up!
In our case, (2,1) eigenvector corresponds to k1=2 and (1,2) eigenvector corresponds to k2=-1, thus we obtain: