Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A\in M(\mathbb F)_{n \times n}$

Prove that the trace of A is minus the coefficient of $\lambda ^{n-1}$ in the characteristic polynomial of A.

I had several ideas to approach this problem - the first one is to develop the characteristic polynomial through the Leibniz or Laplace formula, and from there to show that the contribution to the coefficient of $\lambda ^{n-1}$ is in fact minus the trace of A, but every time i tried it's a dead end. Another approach is to use induction on a similar matrix to ($\lambda I-A$) from an upper triangular form, which has the eigenvalues of A on its diagonal, and of course the same determinant and trace, to show that for every choice of n this statement holds.

I think my proof doesn't hold for all fields, so any thought on the matter will be much appreciated, or an explanation to why this statement is true.

share|cite|improve this question
    
This is basically the same question as math.stackexchange.com/questions/1425082/… – Martin Sleziak Jan 18 at 15:32
    
It is, but im looking for a more detailed answer – Yonatan Izutskiver Jan 18 at 15:35
up vote 4 down vote accepted

The determinant is a sum of (signature-weighted) products of $n$ elements, where no two elements share the same row or column index. From this, it follows, that there is no term with $(n-1)$ terms on the diagonal (if $n-1$ terms of a product are on the diagonal, then the last one must be too, because all other rows and columns are taken). So... the only term that can possibly include a power of $\lambda^{n-1}$ is the product of the main diagonal. Therefore, the $\lambda^{n-1}$ coefficient of $\det A$ equals the $\lambda^{n-1}$ coefficient of $\prod_i (\lambda-A_{ii})$ for which it's easy to show, the coefficient equals $-\sum_i A_{ii}$.

This definition doesn't make assumptions about the field over which the matrix is defined, because the field operations + and * are used directly (with no assumptions about inverses and distribution laws).

share|cite|improve this answer
    
it seems that I lack the algebra, not the theory. I have reached the same conclusion, but failed to show that the $\lambda^{n-1}$ coefficient of $\prod_i (A_{ii}-\lambda)$ equals minus the trace of A. – Yonatan Izutskiver Jan 18 at 15:17
1  
It's similar, again you have all posible products of one coefficient from each term of the product. All those of form $\lambda^{n-1}$ take n-1 terms of $\lambda$ and one other term (there are n combinations, with different diagonal term each time). – orion Jan 18 at 15:40

Suppose the eigenvalues of $A$ are $\lambda_1,\ldots,\lambda_n$. Then the factored form of the characteristic polynomial is $$(x-\lambda_1)(x-\lambda_2)\cdots(x-\lambda_n).$$

Try using this with induction on $n$.

share|cite|improve this answer
    
Sorry, I don't get how it is related to the trace of A... – Yonatan Izutskiver Jan 18 at 15:15
    
Well start with $n=2$. Then we have $(x-\lambda_1)(x-\lambda_2)=x^2-(\lambda_1+\lambda_2)x+\lambda_1\lambda_2$. So we see that the coefficient of the $x$ term is $-(\lambda_1+\lambda_2)=-\operatorname{tr}A$. – Tim Raczkowski Jan 18 at 15:20
    
but $\lambda_1$ and $\lambda_2$ aren't necessarily equal to $a_{1,1}$ and $a_{2,2}$ – Yonatan Izutskiver Jan 18 at 15:23
1  
This proof assumes that the characteristic polynomial can be factored into a product of linear terms. This can be done over an appropriate splitting field but if one doesn't want (or have the tools) to justify the passage to such a splitting field, orion's approach is preferable. – levap Jan 18 at 15:27
    
You can use the fact that $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ and that for any matrix there is a transition matrix $B$ such that $BAB^{-1}$ is an upper triangular matrix whose diagonal entries are the eigenvalues of the matrix. – Tim Raczkowski Jan 18 at 15:33

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.