# Trace and the coefficients of the characteristic polynomial of a matrix

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.

• This is basically the same question as math.stackexchange.com/questions/1425082/… Jan 18, 2016 at 15:32
• It is, but im looking for a more detailed answer Jan 18, 2016 at 15:35

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).

• 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. Jan 18, 2016 at 15:17
• 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). Jan 18, 2016 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 $n$.

• Sorry, I don't get how it is related to the trace of A... Jan 18, 2016 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 $-(\lambda_1+\lambda_2)=-\operatorname{tr}A$. Jan 18, 2016 at 15:20
• but $\lambda_1$ and $\lambda_2$ aren't necessarily equal to $a_{1,1}$ and $a_{2,2}$ Jan 18, 2016 at 15:23
• 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. Jan 18, 2016 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. Jan 18, 2016 at 15:33

You can start by proving that the characteristic polynomial is given by $$p(\lambda) = \lambda^n + (a_{11}-\lambda)...(a_{nn}-\lambda) + q(\lambda)$$ where $$q(\lambda)$$ has degree at most $$n-2$$. Hint: Use induction on $$n$$.

Then it is easy to show that the $$\lambda^{n-1}$$ term has the trace as the coefficient.