This was an exam question that I sort of guessed on.
Find the general solution to $X' = AX$, where $A = \left[\begin{smallmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1\end{smallmatrix}\right]$.
I first found the canonical form of $A$. Then I multiplied that with $X$ to get 3 differential equations and solved those to get the general solution of $A$. My book starts with $A$ in canonical form always so I'm not sure how else to do it if this is incorrect. If I multiply $A$ by $X$ to get 3 differential equations, then they are ones that I don't know how to solve.
$\endgroup$ 01 Answer
$\begingroup$We are asked to find the general solution to:
$$X' = AX = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix}X$$
We can do this many ways, but will use eigenvalues / eigenvectors.
To find the eigenvalues, we set up and solve $|A - \lambda I| = 0$, which yields:
$$-\lambda^3 + \lambda^2 + \lambda -1 \implies \lambda_1 = -1, \lambda_{2,3} = 1$$
To find the eigenvectors, we set up and solve $[A -\lambda_1 I]v_i = 0$. For $\lambda_1 = -1$, we have the row-reduced-echelon-form of:
$$\begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}v_1 = 0$$
This yields the eigenvector:
$$v_1 = (-1, 1, 0)$$
For the eigenvalue $\lambda_{2,3} = 1$, we have a RREF of:
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}v_2 = 0$$
This yields a single independent eigenvector of:
$$v_2 = (0, 0, 1)$$
Now, we need to find another linearly independent generalized eigenvector using $[A-\lambda_3I]v_3 = v_2$, yielding an augmented RREF system of:
$$\left[\begin{array}{ccc|c} 1& 0& 0& \dfrac 12\\ 0& 1& 0& \dfrac 12\\ 0& 0& 0& 0 \end{array}\right]$$
This yields a third eigenvector of:
$$v_3 = \left(\dfrac 12, \dfrac 12, 0\right)$$
Now, we can write the solution of $X' = AX$ as:
$$X(t) = c_1 e^{-t} v_1 + c_2 e^t v_2 + c_3 e^t (v_2 t + v_3)$$
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