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We have a linear operator $X(t): \mathbb{E}^2 \to \mathbb{E}^2$ such that $$\dot X = AX\quad X(0)=\mathbf{1}$$ To be clear $X$ is a $2\times 2$-matrix. An ink spot is contained in $\mathbb{E}^2$ at $t=0$. At which time $t$, will the area of the image of the ink spot under $X(t)$, double for $A=\left(\begin{matrix}-1 & 0\\ 0 & -1\end{matrix}\right)$?


I know the solution is $X(t) = \exp(tA)$. Now the area of a scaling $2\times 2$ matrix is the absolute value of the determinant. So we have

$$|\det(X(t))|= |\det(\exp(tA))|= |\exp(tr(tA))|= \exp(-2t)$$

For $t=0$ we have $1$. So we need to solve $\exp(-2t) = 2$ that is $t= \frac{\ln 2}{-2}$.

Is this solution correct?

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    $\begingroup$ I think you need to be a bit more clear(I'm a bit confused). Is $t \in \mathbb{R}$, or $t \in \mathbb{E}^2$? Is $\dot{X}$ the derivative with respect to $t$? $\endgroup$ – Jeffery Opoku-Mensah Jun 9 '18 at 18:55
  • $\begingroup$ $t\in \mathbb{R}$ and yes, $\dot X$ is the derivative of $X$ wrt $t$. $\endgroup$ – Demmi Li Jun 9 '18 at 18:56
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    $\begingroup$ It seems everything you did is correct. $\endgroup$ – Jeffery Opoku-Mensah Jun 9 '18 at 19:37
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I know the solution is $X(t) = \exp(tA)$. Now the area of a scaling $2\times 2$ matrix is the absolute value of the determinant. So we have

$$|\det(X(t))|= |\det(\exp(tA))|= |\exp(tr(tA))|= \exp(-2t)$$

For $t=0$ we have $1$. So we need to solve $\exp(-2t) = 2$ that is $t= \frac{\ln 2}{-2}$.

Posting my working as an answer to close this question.

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