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$$A = \begin{bmatrix}3&6\\-15&-15\end{bmatrix}$$ has complex eigenvalues $\lambda_{1,2} = a \pm bi$ where $a =$____ and $b = $ ____. The corresponding eigenvectors are $v_{1,2} = c \pm di$ where $c =$ (____ , _____ ) and $d =$ (____ , ___)

So I got the char. poly. eqn $\lambda^2 + 12\lambda + 45 = 0$

Then using the quad. eqn I got $-6 \pm 3i$ which I know is in the form $a \pm bi$ so I was thinking $a = -6$, $b = 3$, but instead they have $b = -3$ , why?

Also, I'm not sure how to obtain the corresponding eigenvectors because we are working with complex eigenvalues now.

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I'll cover the how to find the eigenvectors part.

$$A = \begin{bmatrix}3&6\\-15&-15\end{bmatrix}$$

has eigenvalues $\lambda_{1,2} = -6\pm 3i$.

Now to find the associated eigenvectors, we find the nullspace of $$A-\lambda_{1,2}I = \begin{bmatrix}3-(-6\pm 3i)&6\\-15&-15-(-6\pm 3i)\end{bmatrix} = \begin{bmatrix}9\mp 3i & 6 \\ -15 & -9\mp 3i\end{bmatrix}$$

To find the nullspace I'll put this in REF:

$$\begin{align}\begin{bmatrix}9\mp 3i & 6 \\ -15 & -9\mp 3i\end{bmatrix} &\sim \begin{bmatrix} 5 & 3\pm i \\ 3\mp i & 2\end{bmatrix} \\ &\sim \begin{bmatrix} 5 & 3\pm i \\ 0 & 2-(3\pm i)\frac{-(3\mp i)}{5}\end{bmatrix} \\ &= \begin{bmatrix} 5 & 3\pm i \\ 0 & 0\end{bmatrix}\end{align}$$

Therefore all of the eigenvectors associated with $\lambda_{1,2}$ are of the form $w_{1,2} = \begin{bmatrix}\frac 15(-3\mp i)t \\ t\end{bmatrix}$. Representative eigenvectors are then $$\bbox[5px,border:2px solid red]{v_{1,2} = \begin{bmatrix}-3\mp i \\ 5\end{bmatrix}}$$ which is obtained by setting $t=5$.

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  • $\begingroup$ Why is your $\pm$ signed turned upside down? What's the significance $\endgroup$ – Yusha Aug 2 '16 at 21:20
  • $\begingroup$ It means that $\lambda_1 = -6\oplus 3i$ corresponds to $v_1 = \begin{bmatrix} -3 \ominus i \\ 5\end{bmatrix}$, which I've circled the relevant signs. Likewise $\lambda_2$ and $v_2$ have opposite signs in those places. $\endgroup$ – user137731 Aug 2 '16 at 21:21
  • $\begingroup$ Somehow they are getting $(-1,1)$ and $(-1,2)$ I cant figure out how $\endgroup$ – Yusha Aug 2 '16 at 21:25
  • $\begingroup$ Weird, maybe its an error $\endgroup$ – Yusha Aug 2 '16 at 21:30
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    $\begingroup$ @Yusha Remember that there are an infinite number of vectors in the nullspace of $A-\lambda_{1,2}$. I chose $t=5$ arbitrarily. If you choose $t=1-2i$ you'll get your answer key's solution. As to the reason your book arrived at the answer it did, I'll bet they didn't do the row swap that I did in the first step of my row reduction. So both my and your answer key's solution are correct. $\endgroup$ – user137731 Aug 2 '16 at 21:48
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Note that $\pm (-3)=\mp3=\pm3$. Hence, whether you use $b=3$ or $b=-3$ is irrelevant. To find the corresponding eigenvectors, you need to solve the following:

$$ \begin{bmatrix}3&6\\-15&-15\end{bmatrix}\cdot\begin{bmatrix}x_{1}\\ x_{2}\end{bmatrix}=\lambda_{1,2}\cdot\begin{bmatrix}x_{1}\\ x_{2}\end{bmatrix}. $$

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  • $\begingroup$ I get stuck after I subtract the lambda value of 3 and I get $\begin{bmatrix}9-3i & 6\\-15 & -9-3i\end{bmatrix}$ $\endgroup$ – Yusha Aug 2 '16 at 20:49
  • $\begingroup$ Should I put the matrix in RREF? $\endgroup$ – Yusha Aug 2 '16 at 20:51
  • $\begingroup$ Because If I do $R_2 = 15R_1 + (9-3i)R_2$ this problem begans to get miserable as I will have a quadratic entry.... $\endgroup$ – Yusha Aug 2 '16 at 20:57
  • $\begingroup$ I even row reduced to get $\begin{bmatrix}9-3i&6\\0&0\end{bmatrix}$ and still I'm not getting the correct vectors. this has infinite solutions?? $\endgroup$ – Yusha Aug 2 '16 at 21:06
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The if $b=-3$, $a=-6$, we have $a \pm bi=-6 \pm 3i$, if $b=3, a=-6$ we still have $a\pm bi=-6 \pm 3i$, both answers are correct but if you're submitting into an online homework service like cengage-brain or webassign your professor may have just forgotten to include both cases as acceptable. To find eigenvectors, you need to solve $Av=\lambda v$, or, equivalently $(A-\lambda I)v=0$ (where I denotes the identity matrix. Why are these two systems equivalent?). Nothing really changes with the complex case so long as you know how to do arithmetic in $\mathbb{C}$. To get you started, for an eigenvector corresponding to $\lambda=-6+3i$, we set up: $$ \begin{bmatrix} A-\lambda I \end{bmatrix} v= \begin{bmatrix} 9-3i & 6 \\ -15 & -9-3i \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} =0 $$ Giving the system: $$ \begin{array} ((9-3i)v_1+6v_2=0 \\ -15v_1+(-9-3i)v_2=0 \end{array} $$ Can you take it from here?

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  • $\begingroup$ $v_1 = -6v_2/(9-3i)$, $v_2 = 15v_1/(-9-3i)$? $\endgroup$ – Yusha Aug 2 '16 at 20:41
  • $\begingroup$ I don't get it, don't i need to put the matrix in RREF form? $\endgroup$ – Yusha Aug 2 '16 at 20:47
  • $\begingroup$ Because If I do $R_2 = 15R_1 + (9-3i)R_2$ this problem begans to get miserable as I will have a quadratic entry.... $\endgroup$ – Yusha Aug 2 '16 at 20:57
  • $\begingroup$ I even row reduced to get $\begin{bmatrix}9-3i&6\\0&0\end{bmatrix}$ and still I'm not getting the correct vectors. this has infinite solutions?? $\endgroup$ – Yusha Aug 2 '16 at 21:06
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    $\begingroup$ Yes, there will be infinitely many eigenvectors, but they will all be scalar multiples of one another, your first comment has you basically done, fix $v_1=1$ and solve for $v_2$ in terms of $v_1$, that will give you one eigenvector. Since matrix multiplication is linear, obvious any scalar multiple of this eigenvector will also be an eigenvector since $Av=\lambda v$ implies $A(cv)=c(Av)=c(\lambda v)=\lambda (cv)$. $\endgroup$ – mb- Aug 3 '16 at 2:28

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