0

I was thinking about the following problem:

Let A be a $3\times3$ real valued matrix such that $A^{3}=I$ but $A \neq I$ . Then trace of A must be

(a)0,
(b)1,
(c)-1,
(d)3

I have went through the answer which says that $x=1,w,w^{2}$ are it's eigenvalues and hence $trace(A)=0$(A be a $3\times3$ real valued matrix such that $A^{3}=I$ but $A \neq I$ .Then trace(A)=?). But, since we have considered matrix over real field ,how can i have complex eigenvalues at all.

my confusion arises from an example in hoffman's text which is

$ A =$$ \begin{pmatrix} 0 & -1 \\ 1 &0 \\ \end{pmatrix} $$ $

then,$ch_A(x)$=$x^2+1$,which has no real roots,so,A has no eigenvalues.So, how in above problem we have complex eigenvalues??

thanks for your time in advance...

ajeet
  • 25
  • 4
  • 2
    Your example has no real eigenvalues. However every matrix with real entries has complex eigenvalues. This follows from fundamental theorem of algebra. – Sahiba Arora Jun 08 '17 at 13:37
  • @sahiba arora:then,how to find eigenvalues of the problem,i see that $ch_A(x)=x^3-1$ but how do we get eigenvalues from here??because now this equation has only 1 as real root and since matrix is considered over $\Bbb R$ so we cannot solve the equation for complex roots...so how – ajeet Jun 08 '17 at 13:54
  • When the matrix is considered over $\mathbb{R}$, it means the entries are real. The matrix definitely has complex eigenvalues. It does not mean you ignore the complex eigenvalues. – Sahiba Arora Jun 08 '17 at 14:13
  • @sahiba arora:this means that real matrix can have complex eigenvalues...but roots of characteristic polynomial of this real matrix gives us only the roots which belong to field $\Bbb R$.for finding complex roots,we have to think in other way...like you did in your answer – ajeet Jun 08 '17 at 14:20
  • Real matrix always has complex eigenvalues. Eigenvalues are just roots of characteristic polynomial. The characteristic polynomial will give you all roots including the ones that don't belong to $\mathbb{R}$. – Sahiba Arora Jun 08 '17 at 14:23
  • @sahiba arora:then as given in example above,why does hoffman book saya that A has no eigenvalues(I think that he says it because in definition of eigenvalue itself ,eigenvalues are considered only over the field of matrix) – ajeet Jun 08 '17 at 14:34
  • I checked the book. Prior to the example, Hoffman defines characteristic value over the field $\mathbb{F}$. In the example he is trying to illustrate that eigenvalues may not exist in the field $\mathbb{F}$. However, the roots will certainly exist in the algebraic closure of $\mathbb{F}$ which in our case is $\mathbb{C}$. However if you are working on some problem related to eigenvalues you certainly need all the eigenvalues of the matrix to give you the information you need. – Sahiba Arora Jun 08 '17 at 14:47
  • @sahiba arora:ok...thanks a lot for your clarification...but, i strictly think if that is the case then $\lambda\in\Bbb F$ should not be mentioned in definition... – ajeet Jun 08 '17 at 14:55

2 Answers2

1

Another way to see this: if the matrix $A$ has eigenvalues $\lambda_1, \lambda_2, \lambda_3$, then the eigenvalues of $A^3$ are $\lambda_1^3, \lambda_2^3$, and $\lambda_3^3$. But the eigenvalues of $A^3 = I$ are all 1 (with three-fold degeneracy), which means that $\lambda_1^3 = \lambda_2^3 = \lambda_3^3 = 1$. Thus, the $\lambda_i$ are all cube roots of unity. Moreover, since $A$ is real-valued, the trace of $A$ is real, and $\mathrm{tr}(A) = \lambda_1 + \lambda_2 + \lambda_3$. There are only two ways for this quantity to be real: either $\lambda_1 = \lambda_2 = \lambda_3 = 1$ (which implies $A = I$, a contradiction) or $\lambda_1 = 1$, $\lambda_2 = e^{2 \pi i/3}$, and $\lambda_3 = e^{4 \pi i/3}$. In the latter case, $\lambda_1 + \lambda_2 + \lambda_3 = 0$.

Michael Seifert
  • 9,477
0

$$ch_A(x) = x^3 -1=(x-1)(x^2+x+1)$$

Therefore roots of $ch_A(x)$ are $1$, $\frac{-1+\sqrt{3}i}{2}$ and $\frac{-1-\sqrt{3}i}{2}$

Let $\omega =\frac{-1+\sqrt{3}i}{2}$. Then $\omega ^2 = \frac{-1+\sqrt{3}i}{2}$.

Clearly, $\omega$ satisfies $1+ \omega + \omega^2 =0$.

Sahiba Arora
  • 10,847