Linear homogenous system with repeated eigenvalues

[psie]

Homework Statement

Solve the IVP $x'=Ax, x(0)=x_0$, where $A=\begin{pmatrix} 1&1&-1\\ 0&2&-2\\-1&1&-1\end{pmatrix}$ and $x_0=\begin{pmatrix} -1 \\ 1\\ 0\end{pmatrix}$.

Relevant Equations

Determinants, Gaussian elimination, generalized eigenvectors, matrix exponential, etc.

I've solved this problem using a fairly involved technique, where I compute the matrix $e^{tA}$ (the fundamental matrix of the system) with a method derived from the Cayley-Hamilton's theorem. It is a cool method that I believe always works, but it can be a lot of work sometimes. It involves finding a polynomial function, which you then evaluate at the matrix $A$. Once you've found $e^{tA}$, the general solution to the system is $e^{tA}x_0$

Anyway, I've been presented another solution and I'm not really sure what formula they used in the end. To be honest, their whole approach is very new to me and I'd be grateful for any references where this is explained in more detail.

So the eigenvalues of $A$ are $2$ and $0$, the latter with multiplicity $2$. The solution goes then as follows:

The generalized eigenvectors for $\lambda=0$ are given by the null space to $$A^2=\begin{pmatrix} 2&2&-2 \\ 2&2&-2\\0&0&0\end{pmatrix}$$ which has a basis $(0,1,1)$ and $(1,0,1)$. The eigenvalue for $\lambda=2$ are given by the null space to $$A-2I=\begin{pmatrix} -1&1&-1 \\ 0&0&-2\\-1&1&-3\end{pmatrix}$$ which gives the eigenvector $(1,1,0)$. The general solution is \begin{align} x(t)&=(I+tA)(c_1(0,1,1)+c_2(1,0,1))+c_3e^{2t}(1,1,0) \tag1 \\&=c_1(0,1,1)+c_2((1,0,1)-2t(0,1,1))+c_3e^{2t}(1,1,0) \tag2\end{align}

The initial value $(-1,1,0)=(0,1,1)-(1,0,1)$ is a generalized eigenvector for $\lambda=0$ so we get the solution $x(t)=(I+tA)(-1,1,0)=(-1,1,0)+t(0,2,2)$.

1. What formula are they using in $(1)$, i.e. where does the $(I+tA)$ come from?
2. The initial value is a linear combination of generalized eigenvectors. I do not understand how they conclude $x(t)=(I+tA)(-1,1,0)$?

Grateful if someone could answer both questions and possibly refer to some text where this method is explained in more detail.

 

[pasmith]

1) If A has an eigenvalue \lambda with multiplicity 2, then there exist non-zero vectors u and v such that \begin{split}<br /> (A - \lambda I)v &amp;= u, \\<br /> (A - \lambda I)u &amp;= 0.\end{split} It follows that if x = a(t)u + b(t)v and \dot x = Ax then <br /> \begin{split} \dot a u + \dot b v &amp;= A(au + bv) \\<br /> &amp;= \lambda a u + b (\lambda v + u) \end{split} so that \begin{split}<br /> \dot a &amp;= \lambda a + b \\<br /> \dot b &amp;= \lambda b \end{split} and \begin{split}<br /> x(t) &amp;= a_0 e^{\lambda t} u + b_0 e^{\lambda t} (tu + v) \\<br /> &amp;= a_0 e^{\lambda t} u + b_0 e^{\lambda t} (t(A - \lambda I) + I)v. \end{split} Since (A - \lambda I)u = 0 we can add a_0e^{\lambda t}t(A - \lambda I)u = 0 to the right hand side to get <br /> \begin{split}<br /> x(t) &amp;= a_0 e^{\lambda t} (t(A - \lambda I) + I)u + b_0 e^{\lambda t} (t(A - \lambda I) + I)v \\<br /> &amp;= e^{\lambda t}(t(A - \lambda I) + I)(a_0 u + b_0 v). \end{split}

2) You need to solve <br /> (-1, 1, 0) = x(0) = c_1(0,1,1) + c_2(1,0,1) + c_3 (1,1,0). It happens that (c_1, c_2, c_3) = (1, -1, 0).

 

[pasmith]

A better explanation is that here \mathbb{R}^3 = \ker A^2 \oplus \ker (A - 2I), so that any x \in \mathbb{R}^3 can be written uniquely as the sum of a vector v \in \ker A^2 and a vector u \in \ker (A - \lambda I).

For v \in \ker A^2 we have A^2 v = 0 so that e^{tA}v = (tA + I)v and for u \in \ker (A - 2I) we have Au = 2u so that e^{tA}u = e^{2t}u. Hence \begin{split}<br /> e^{tA}x &amp;= e^{tA}v + e^{tA}u \\<br /> &amp;= (tA + I)v + e^{2t}u.\end{split}