See the attachment
update: To find a basis to quasi-diagonalize A, we need to find pick e3, such that, the minimal polynomial f satisfying f(A)e3 = 0, is the f(x)=(x-1)^3 (the characteristic polynomial of a Jordan block)(this holds for a “general enough” e3). In this case, if you take e2, e1 as in the answer, then you find A is quasi-diagonalized to a Jordan block. And e1 is the eigenvector, as you can see from the matrix.
e3 can be other vectors, as long as the condition is satisfied.
update2：ok, let’s try another way.
First we know the characteristic polynomial of A is (x-1)^3.
Then we calculate the eigenvector, we find there is only one eigenvector e1=(4, -1, 0) up to multiplying by a non zero scalar. So the Jordan canonical form must be as in the answer.
Then we need to find e2, e3, such that (A-I)e2=e1, (A-I)e3=e2. If so, then under the basis {e1, e2, e3}, A will quasi-diagonalized to the Jordan block.
We need to solve (A-I)e2=e1. In general, to solve Ax=b, we need to do row operations to make A into the row-reduced echelon form. This row operations amount to mutiply A by an invertible matrix P on left. Now it suffices to solve PAx=rref(A)x=Pb. But rref matrix is easy so you can find a solution immediately.
So we find e2, similarly e3, then solve the remain of the problem as in the attachment.
update3: your thoughts are right, but dy/dt = Jy is also not wrong.
If you wants to directly solve the equation dx/dt = Ax + f , then you can use dy/dt = Jy + g. Solve y and find x = Py.
If you wants to first solve the homogeneous equation dx/dt = Ax , then use variation of constants to solve dx/dt = Ax + f. Then in this process you only need to consider dy/dt = Jy, in correspondence to dx/dt = Ax.
Two ways are equivalent.
update4: “y” and “C” are just names. So if you rename them as y1 = C1, y2 = C2, y3 = C3, then you find homogeneous solution = C1 * Sol1 + C2 * Sol2 + C3 * Sol3.
To determine a solution of an ODE, you need not only the equation, but also initial conditions (like x(0) = …, x\'(0)= …). Here, y0 = (y1, y2, y3) = y(0) is the initial condition of dy/dt = Jy. y1, y2, y3 can be any constant. The undetermined constants “y1, y2, y3” and “C1, C2, C3” just represent lack of knowledge of some initial conditions.
Reminder: general solution = a particular solution + homogeneous solution
update5: maybe my notation makes you confused. y1, y2, y3 are just undetermined constants, they are not solutions of the equation. They shouldn\’t be solved.
If you take y1=1, y2=0, y3=0. Then you find a solution to the homogeneous equation f1(t) = e^t (8, -2, 0)
If you take y1=0, y2=1, y3=0. Then you find a solution to the homogeneous equation f2(t) = e^t (-2+8t, -1-2t, -4)
If you take y1=0, y2=0, y3=1. Then you find a solution to the homogeneous equation f3(t) = e^t (1-2t+4t^2, -t-t^2, -4t)
If you take y1 = C1, y2 = C2, y3 = C3. Then you find the general solution to the homogeneous equation C1 f1(t) + C2 f2(t) + C3 f3(t).
We have found a particular solution x0(t)=e^t (t-8t^3/3, 3t^2/2+2t^3/3, 2t+4t^2)
So the solution to the original equation is x(t) = x0(t) + C1 f1(t) + C2 f2(t) + C3 f3(t).
