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Let $y(t)=\begin{pmatrix} y_1(t)\\ y_2(t) \end{pmatrix}$ satisfy $\dfrac {dy}{dt}=Ay; t>0, y(0)=\begin{pmatrix} 0\\ 1 \end{pmatrix}$ where $A$ is a $2 \times 2$ constant matrix with real entries satisfying trace $A=0$ and det $A >0.$ Then which of the following options are true?
1. $y_1(t)$ and $y_2(t)$ both are oscillating functions of $t$.
2. monotonically decreasing functions of $t$
3. monotonically increasing functions of $t$
4. constant functions of $t$.

My Attempt: We take $A=\begin{pmatrix} 2 &-6 \\ 1& -2 \end{pmatrix}$ so that trace $A=0$ and det $A >0.$ Now $\dfrac {d}{dt} \begin{pmatrix} y_1(t)\\ y_2(t) \end{pmatrix}=\begin{pmatrix} 2 &-6 \\ 1& -2 \end{pmatrix}\begin{pmatrix} y_1(t)\\ y_2(t) \end{pmatrix} \implies \begin{pmatrix} y_1'(t)\\ y_2'(t) \end{pmatrix}=\begin{pmatrix} 2y_1-6y_2\\ y_1-2y_2 \end{pmatrix}$. Now $y_1'(t)=2y_1-6y_2 \implies y_1''(t)=2y_1'-6y_2'=2y_1'-6(y_1-2y_2') \implies y_1''+2y_1=0.$ Hence $y(x)=c_1 \cos \sqrt 2x+c_2 \sin \sqrt 2x$ which is an oscillating function. Here I am not sure whether I have to use the condition $y(0)=\begin{pmatrix} 0\\ 1 \end{pmatrix}$ ,which I did not use at all.

Am I going in the right direction? If not can someone point me in the right direction? Any better approach of tackling the problem will be appreciated. Thanks in advance for your time.

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I don't think, that the task of the proof is to show, that there exist an example. I think, you need to describe what does that mean that function is oscilating and proove, that it requires the matrix to have zero trace and determinant. –  V-X Apr 2 '13 at 16:34
    
I also don't think, that det A of your example is zero $2(-2) - (-6)1 = 2$ –  V-X Apr 2 '13 at 16:35
    
You're sort of correct, but just testing a single example isn't enough for this problem. It asks you what properties hold for any matrix satisfying the given constraints. –  kahen Apr 2 '13 at 17:07

1 Answer 1

up vote 1 down vote accepted

Yes you are, here are some hints. An initial value problem of the type

$\frac{dy}{dt}=Ay,\quad\quad y(0)=y_0\in\mathbb{R}^n$

where $A$ is an $n\times n$ real diagonalisable matrix has a unique solution given by

$y(t)=\sum_{i=1}^n c_iv_ie^{\lambda_i t}\quad\quad\quad(*)$

where $\lambda_i$ denote the eigenvalues of $A$, $v_i$ denotes the eigenvector corresponding to $\lambda_i$ and $c_i\in\mathbb{C}$ denote constants picked such that $y(0)=y_0$. In addition, note the identities

$trA=\lambda_1+\lambda_2+\dots+\lambda_n$ and $\quad detA=\lambda_1\lambda_2\dots\lambda_n$,

what do they imply about the eigenvalues of your matrix? Let me know if you would like me to elaborate any further.


EDIT: $A$ is $2\times 2$, so at most it has $2$ distinct eigenvalues, $\lambda_1$ and $\lambda_2$. Since trace$(A)=0$ and trace$(A)=\lambda_1+\lambda_2$, we must have that $\lambda_2=-\lambda_1$.

Next, $\det(A)=\lambda_1\lambda_2=-\lambda_1^2>0$. The inequality is only satisfied if $\lambda_1$ is an imaginary number, that is $\lambda_1=\alpha i$ where $\alpha\neq0$ is some real number.

Since $A$ has two distinct eigenvalues, it is diagonalisable. Which implies that $y$ must be of the form of $(*)$. That is,

$$y(t)=c_1v_1e^{\lambda_1 t}+c_2v_2e^{\lambda_2 t}=c_1v_1e^{\alpha i t}+c_2v_2e^{-\alpha i t}=c_1v_1[\cos(\alpha t)+i \sin(\alpha t)]+c_2v_2[\cos(\alpha t)-i \sin(\alpha t)].$$

In other words, $y_1$ and $y_2$ are linear combinations of periodic functions, and thus must periodic themselves (unless they exactly cancel out leaving $y\equiv 0$, however you can rule this out since $y(0)=[0$ $1]^T$).

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tr $A=0$ means that eigenvalues must be of opposite sign whereas det $A>0$ means eigenvalues must be of same sign(either both positive or both negative),since det $A=\lambda_1 \lambda_2.$ –  learner Apr 2 '13 at 17:18
    
Eigenvalues can be complex numbers (in particular, imaginary) –  jkn Apr 2 '13 at 17:27
    
@learner see the edit - hopefully it helps. –  jkn May 29 '13 at 10:30
    
@jkn- I hope you could clear my one doubt- How do we know if a matrix is not diagonalizable? –  Ramit Aug 13 '13 at 9:56
    
@Ramit A matrix is diagonalisable if and only if you can express any vector as a linear combination of the matrix's eigenvectors (that is, if the set of eigenvectors span the whole space). If so, for any $x\in\mathbb{R}^n$ we can write $x=Vy\Rightarrow y=V^{-1}x$, thus $Ax=AVy=V\Lambda y=V\Lambda V^{-1}x$. $V$ and $\Lambda$ denote the matrix of eigenvectors (as columns) and eigenvalues (on the diagonal) of $A$, respectively. –  jkn Aug 27 '13 at 10:23

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