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$A=\begin{bmatrix} a & b\\ b & c \end{bmatrix}$

Find any $(a,b,c) \in \mathbb{C}^3$ , $a \neq 0$ , $b \neq 0$ , $c \neq 0$ , so that eigenvalues of A are $\lambda_1=\lambda_2=1$

$det(A-\lambda I)=\begin{bmatrix} a-\lambda & b\\ b & c- \lambda \end{bmatrix}=(a-\lambda)(c-\lambda)-b^2=0$

$\Longrightarrow$ $\lambda^2-\lambda(a+c)+ac-b^2=0$

$\lambda=\frac{a+c \pm \sqrt{a^2+4b^2-2ac+c^2}}{2}$ $\Longrightarrow$

$a^2+4b^2-2ac+c^2=0$ and $\frac{a+c}{2}=1$

(discriminant is 0 because I want a double root, and (a+c)/2=1 because I want it to be 1)

but I can't find any solution, is there something wrong?





$a= 1 \pm b \cdot i $

Whatever value I give to a, it won't work...

share|cite|improve this question
\Longrightarrow to get $\Longrightarrow$. \cdot to get $\cdot$ or \times to get $\times$. – Git Gud Jan 23 '13 at 21:00
thank you! i will replate * with dot – Cristi Jan 23 '13 at 21:02
@all $\implies$ (\implies) is a little easier to type. – rschwieb Jan 23 '13 at 21:07
@Cristi Did you want any matrix with eigenvalues all equal to $1$? If that's the case, the identity matrix is a simple example. If that's not the case, I don't see how rschwieb's answer solves the problem of finding all the matrixes that you're looking for, he just gives one instance of such matrices. – Git Gud Jan 23 '13 at 21:15
I found all the solutions: $(a,\pm i(a-1),2-a) \in \mathbb{C}^3$ – Cristi Jan 23 '13 at 21:26
up vote 4 down vote accepted

You are very close, everything is okay up to the step: $4a^2-8a+4b^2+4=0$.

This reduces to $a^2-2a+b^2+1=0$

Now, solve for

$$b = \pm \sqrt{-(a-1)^{2}} = \pm i(a-1)$$

So, you are free to choose $a's$.


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thanks! it woks now – Cristi Jan 23 '13 at 21:23
Excellent +1; it "took"! – amWhy May 7 '13 at 0:17

Your work looks OK up to $4a^2-8a+4b^2+4=0$.

This reduces to $a^2-2a+b^2+1=0$, and you can try a few values for $a$ and $b$. I tried $a=3$, which yielded $c=-1$ and $b=2i$. The resulting matrix has the eigenvalues you want.

share|cite|improve this answer
$4a^2-8a+4b^2+4=0$ is the same thing as $a^2-2a+b^2+1=0$ , now I see it works, after many edits... thank you! – Cristi Jan 23 '13 at 21:19

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