Science:Math Exam Resources/Courses/MATH152/April 2017/Question B 04 (b)

MATH152 April 2017

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Question B 04 (b)
A model of love affairs is described by the following system of differential equations ${\begin{aligned}R'&=R+4J,\\J'&=-2R-3J,\end{aligned}}$ where $R(t)$ and $J(t)$ are, respectively, Romeo’s and Juliet’s amount of love for each other at time $t$. (b) The eigenvalues of $A$ are $\lambda =-1\pm 2i$. Find the eigenvectors of the matrix with these eigenvalues.

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Hint
Recall the definition of eigenvectors. $v\neq 0$ is an eigenvector of a matrix $A$ with corresponding eigenvalue $\lambda$ if $Av=\lambda v$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We know from part a) that $A={\begin{bmatrix}\quad \!1&\quad \!4\\-2&-3\\\end{bmatrix}}$. The eigenvector $v={\begin{bmatrix}R\\J\\\end{bmatrix}}$ with eigenvalue $\lambda =-1+2i$ must satisfy the eigenvector equation $Av=\lambda v$. So, we have ${\begin{bmatrix}\quad \!1&\quad \!4\\-2&-3\\\end{bmatrix}}{\begin{bmatrix}R\\J\\\end{bmatrix}}=(-1+2i){\begin{bmatrix}R\\J\\\end{bmatrix}}$, which yields the two equations: $R+4J=-R+2iR$ $-2R-3J=-J+2iJ$ Using the first of these equations, we obtain the constraint: $J={\frac {(-1+i)}{2}}R$. The second equation gives the constraint: $(-1-i)J=R$, but this is equivalent to the previous constraint since ${\frac {1}{-1-i}}={\frac {-1+i}{2}}$. Since any nonzero multiple of an eigenvector is also an eigenvector with the same eigenvalue, we can pick $R=1$ to get the eigenvector, $v={\begin{bmatrix}1\\{\frac {-1+i}{2}}\\\end{bmatrix}}$ with eigenvalue $\lambda =-1+2i$ We do the same for the second eigenvalue $\lambda =-1-2i$ to get the eigenvector equation: ${\begin{bmatrix}\quad \!1&\quad \!4\\-2&-3\\\end{bmatrix}}{\begin{bmatrix}R\\J\\\end{bmatrix}}=(-1-2i){\begin{bmatrix}R\\J\\\end{bmatrix}}$. This time, we get the two equations: $R+4J=-R-2iR$ $-2R-3J=-J-2iJ$ which leads to the constraint $J={\frac {(-1-i)}{2}}R$. Again, choosing $R=1$ we get the eigenvector $v={\begin{bmatrix}1\\{\frac {-1-i}{2}}\\\end{bmatrix}}$ associated to eigenvalue $\lambda =-1-2i$. Answer: $\color {blue}{\begin{bmatrix}1\\{\frac {-1+i}{2}}\\\end{bmatrix}}{\text{with eigenvalue }}\lambda =-1+2i,\quad {\begin{bmatrix}1\\{\frac {-1-i}{2}}\\\end{bmatrix}}{\text{with eigenvalue }}\lambda =-1-2i$