Science:Math Exam Resources/Courses/MATH152/April 2012/Question 08 (b)

MATH152 April 2012

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Question 08 (b)
Consider the 2 x 2 system of differential equations $\mathbf {x} '(t)={\begin{bmatrix}-1&-2\\2&-1\end{bmatrix}}\mathbf {x} (t)$ where $\mathbf {x} (t)={\begin{bmatrix}x_{1}(t)\\x_{2}(t)\end{bmatrix}}$ is the vector of unknown functions. Find the corresponding eigenvectors.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The eigenvectors are the non-trivial vectors that satisfy $\displaystyle {}(A-\lambda {}I)\mathbf {x} =\mathbf {0} .$ Setting the determinant to be zero (when finding the eigenvalues) is what makes such vectors possible.

Hint 2
Calculate vectors $\displaystyle x$ and $\displaystyle y$ which satisfy ${\begin{pmatrix}-1-\lambda _{1}&-2\\2&-1-\lambda _{1}\end{pmatrix}}x=0$ and ${\begin{pmatrix}-1-\lambda _{2}&-2\\2&-1-\lambda _{2}\end{pmatrix}}y=0$ These correspond to the two eigenvalues from part (a).

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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 first calculate the eigenvector $\displaystyle x$ of the eigenvalue $\displaystyle -1-2i$, $0={\begin{pmatrix}-1-(-1-2i)&-2\\2&-1-(-1-2i)\end{pmatrix}}x={\begin{pmatrix}2i&-2\\2&2i\end{pmatrix}}x={\begin{pmatrix}2ix_{1}-2x_{2}\\2x_{1}+2ix_{2}\end{pmatrix}}=0$ Considering the first line, we obtain ${\begin{aligned}\displaystyle 2ix_{1}-2x_{2}&=0\\x_{2}&=ix_{1}\end{aligned}}$ Notice that there is a free variable which is common when finding eigenvectors. Therefore, we can choose $\displaystyle x_{1}=1$, then $\displaystyle x_{2}=i$ and have the eigenvector $\displaystyle x={\begin{pmatrix}1\\i\end{pmatrix}}$. For the second eigenvector $\displaystyle y$ of the eigenvalue $\displaystyle -1+2i$, we calculate $0={\begin{pmatrix}-2i&-2\\2&-2i\end{pmatrix}}y={\begin{pmatrix}-2iy_{1}-2y_{2}\\2y_{1}-2iy_{2}\end{pmatrix}}=0$ Considering the first line, we obtain ${\begin{aligned}\displaystyle -2iy_{1}-2y_{2}&=0\\y_{2}&=-iy_{1}\end{aligned}}$ There is a free choice like before and we choose $\displaystyle y_{1}=1$, then $\displaystyle y_{2}=-i$ and have the eigenvector $\displaystyle y={\begin{pmatrix}1\\-i\end{pmatrix}}$.