Initial Value Problem with Repeated Eigenvalues Given the matrix
$$
A=\left(\begin{array}{ccc}
1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)
$$
For $X'= AX.\quad$
$X\left(0\right)=\left(\begin{array}{r}1 \\ 0 \\ -2\end{array}\right)\,.\quad$
What is the solution ?.
 A: We find the eigenvalues and eigenvectors as:


*

*$\lambda_{1,2,3} = 1$, which gives one eigenvector and two generalized eigenvectors as:

*$v_1 = (0,1,0), v_2 = (1,0,0), v_3 = (0,0,1)$


We can write the general solution as:
$$X(t) = e^t\left(c_1v_1 + c_2(t v_1 + v_2) + c_3\left(\frac{t^2}{2!}v_1 + t v_2 + v_3\right)\right)$$
Using the initial condition, $X(0)$, we find:
$c_1 = 0, c_2 = 1, c_3 = -2$
Our final solution is then:


*

*$x(t) = e^t(1 - 2t)$

*$y(t) = e^t(t - t^2)$

*$z(t) = e^t(-2)$


Note, there are many ways to do these types of problems from the matrix exponential, fundamental matrix, set of linear equations...
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$$
X\left(t\right) = {\rm e}^{At}\,X\left(0\right)
=
\expo{t}{\rm e}^{\pars{A - 1}t}\,X\left(0\right)
=
\expo{t}\bracks{1 + \pars{A - 1}\,t + \half\,\pars{A - 1}^{2}\,t^{2}}X\left(0\right)
$$
since $\pars{A - 1}^{3} = 0$ where
$$
A - 1 = \pars{\begin{array}{ccc}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array}}
\qquad\mbox{and}\qquad
\pars{A - 1}^{2}
= \pars{\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}}
$$
such that:
$$
1 + \pars{A - 1}t + \half\,\pars{A - 1}^{2}t^{2}
=
\pars{\begin{array}{ccc}
1 & 0 & t \\[1mm] t & 1 & \half\,t^{2} \\[1mm] 0 & 0 & 1 \end{array}}
$$
Then,
$$
X\pars{t}
=
\expo{t}\pars{\begin{array}{ccc}
1 & 0 & t \\[1mm] t & 1 & \half\,t^{2} \\[1mm] 0 & 0 & 1 \end{array}}
\pars{\begin{array}{c}1 \\[3mm] 0 \\[3mm] -2  \end{array}}
=
\expo{t}\pars{\begin{array}{c}1 - 2t\\[3mm] t - t^{2} \\[3mm] -2  \end{array}}
\quad\imp\quad
\left\lbrace\color{#0000ff}{\large%
\begin{array}{rcr}
{\rm x}\pars{t} & = & \pars{1 - 2t}\expo{t}
\\
{\rm y}\pars{t} & = & t\pars{1 - t}\expo{t}
\\
{\rm z}\pars{t} & = & -2\expo{t}
\end{array}}\right.
$$
