Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A$ be an $n\times n$ matrix with entries in an arbitrary field $k$. Is the characteristic polynomial $\det(tI_n-A)$ dependent only on the trace and determinant of $A$?

share|improve this question
add comment

2 Answers 2

up vote 1 down vote accepted

No, look at $\left[\matrix{1&0&0\cr0&1&0\cr0&0&0 }\right]$ and $\left[\matrix{2&0&0\cr0&0&0\cr0&0&0 }\right]$. Both have determinant 0 and trace 2 but they do not have the same eigenvalues. Similar examples show it's not true for any $n>3$.

share|improve this answer
add comment

It's not true for $n\geq 3$. Take the matrix $A:=\pmatrix{0&1&0\\\ 0&0&1\\\ a_0&a_1&a_2}$ (companion matrix). Its characteristic polynomial is $-X^3+a_2X^2+a_1X+a_0$ and its determinant and trace are respectively $a_0$ and $a_2$. But the characteristic polynomial depends on $a_1$, which is independent of the trace and the determinant.

share|improve this answer
    
Maybe I'm wrong but shouldn't you be saying for $n \ge 4$ since, for a $3 \times 3$ case, the characteristics polynomial still stays as a function of trace and determinant. Wiki stands as reference. –  Inquest Mar 9 '12 at 18:42
    
It's the trace of $A^2$ not the trace of $A$ (I guess the OP wanted an answer with only the trace of $A$ and the determinant of $A$, but maybe I misunderstood the problem). –  Davide Giraudo Mar 9 '12 at 18:44
    
or maybe I did..... –  Inquest Mar 9 '12 at 18:48
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.