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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.

My attempts: I take A to be $$\begin{pmatrix} x &0 &0 \\ 0& x &0 \\ 0&0 & x \end{pmatrix}.$$ Now we see $A^{3}=I$ gives $x^{3}=1$ which gives $x=1,w,w^{2}$ where $w$ being the cube root of unity. Thus we see that trace of A is $1+w+w^{2}=0$. So ,i think that (a) is the right choice.Am i going in the right direction? Please give your valuable opinion. Thanks in advance for your time.

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Your argument doesn't work in general. For instance, the diagonal matrix with entries $1, \omega, \omega$ also satisfies $A^3=I$, but its trace is non-zero. You have to use the fact that $A$ is a real matrix. –  TonyK Dec 11 '12 at 9:25
2  
Hint 1: The Trace of a matrix is the sum of its eigenvalues. Hint 2: If $A$ is a real matrix, then its complex eigenvalues come in pairs $(a+bi, a-bi)$. –  TonyK Dec 11 '12 at 9:30

1 Answer 1

up vote 3 down vote accepted

Since $A^3=I \Longrightarrow$ the minimal polynomial of $A$ divides $x^3-1$. Thus the eigenvalues of $A$ are in $\{1,\omega,\omega^2\}$ ($\omega$ is a primitive third root of unity).
Since $A$ is real if $\omega$ is an eigenvalue of $A$ then $\bar{\omega}=\omega^2$ is an eigenvalue of $A$. The same if $\omega^2$ is an eigenvalue.
Therefore we have only two possibilities for the eigenvalues of $A$. Either $1,1,1$ or $1,\omega,\omega^2$. If the eigenvalues are $1,1,1$ then the minimal polynomial of $A$ is $x-1$, thus $A=I$ (why?).
Therefore if $A\neq I$ the eigenvalues are $1,\omega,\omega^2$ and the trace is $0$.

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1) This is a homework problem so don't just give the solutions away; 2) your answer is essentially the same as mine. –  user38268 Dec 11 '12 at 9:54
    
@BenjaLim It seems that he knows that if $1,\omega,\omega^2$ are the eigenvalues then the trace is $0$. Indeed your answer is the same as mine... –  P.. Dec 11 '12 at 9:58
    
And both your answers are the same as my comment :-) –  TonyK Dec 11 '12 at 11:41
    
@TonyK: Yes they are :-) –  P.. Dec 11 '12 at 12:33

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