Science:Math Exam Resources/Courses/MATH152/April 2017/Question B 04 (b)/Solution 1

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: $\color {blue}{\begin{bmatrix}1\\{\frac {-1+i}{2}}\\\end{bmatrix}}{\text{with eigenvalue }}\lambda =-1+2i,\quad {\begin{bmatrix}1\\{\frac {-1-i}{2}}\\\end{bmatrix}}{\text{with eigenvalue }}\lambda =-1-2i$