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I have looked extensively for a proof on the internet but all of them were too obscure. I would appreciate if someone could lay out a simple proof for this important result. Thank you.

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up vote 57 down vote accepted

These answers require way too much machinery. By definition, the characteristic polynomial of an $n\times n$ matrix $A$ is given by $$p(t) = \det(A-tI) = (-1)^n \big(t^n - (\text{tr} A) \,t^{n-1} + \dots + (-1)^n \det A\big)\,.$$ On the other hand, $p(t) = (-1)^n(t-\lambda_1)\dots (t-\lambda_n)$, where the $\lambda_j$ are the eigenvalues of $A$. So, comparing coefficients, we have $\text{tr}A = \lambda_1 + \dots + \lambda_n$.

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I assume you mean the first equation. Use your favorite definition for computing determinants. How do we arrive at a coefficient of $t^{n-1}$? It has to come from multiplying all the diagonal elements of $A-tI$, using the $-t$ from $n-1$ of the factors and the $a_{ii}$ from the remaining factor, and then adding such terms. – Ted Shifrin Oct 30 '13 at 22:29
Write out the $2\times 2$ and $3\times 3$ cases completely. What formula do you get for $\det(A-tI)$? – Ted Shifrin Oct 30 '13 at 23:06
What is $b_{n-1}$ in terms of the entries of $A$? Please do as I asked, and do the cases $n=2,3$ by brute force. – Ted Shifrin Oct 30 '13 at 23:15
@Ioannis Well, if you're not able to see how you arrive to coefficient in the general case from writing down the cases $n=2,3$, then the only possiblity is to give you a very technical proof by definition. Is that what you seek? – yo' Oct 30 '13 at 23:36
I don't get how you got the characteristic polynomial in terms of the trace like that. I'm pretty sure that's not "by definition". – Mehrdad Mar 4 '15 at 21:55

I'll try to show it another way. We know that if we have a polynomial $x^n+b_{n-1} x^{n-1} + \dots +b_1 x+ b_0$, then $(-1)^{n-1} b_{n-1}$ is the sum of the roots of this polynomial. (So-called Vieta's formulas) In our case, the polynomial is $\det(tI-A)$ and we have $(-1)^{n-1} b_{n-1}=\lambda_1+\lambda_2+\dots+\lambda_n$.

$\def\S{\mathcal{S}_n}$ Let $\S$ denote all the permutations of the set $\{1,2,\dots,n\}$. Then by definition $$ \det M = \sum_{\pi\in\S} m_{1,\pi(1)} m_{2,\pi(2)} \dots m_{n,\pi(n)} \operatorname{sgn}\pi, $$ where $\operatorname{sgn}\pi$ is either $+1$ or $-1$ and it is $+1$ for the identity permutation (we don't need to know more now).

Consider $M=tI-A$. To get the power $t^{n-1}$ for a permutation, we need this permutation to choose at least $n-1$ diagonal elements, i.e., to have $\pi(i)=i$ for at least $n-1$ values of $i$. However, once you know the value of a permuation on $n-1$ inputs, you know the last one as well. This means, that to get the coefficient of $t^{n-1}$, we need to consider only the identity permutation.

So far we got that $b_{n-1}$ is the coefficient of $t^{n-1}$ in $(t-a_{1,1})(t-a_{2,2})\dots(t-a_{n,n})$ (this is the term of the sum above corresponding to the identity permutation). Therefore $(-1)^{n-1}b_{n-1} = a_{1,1}+a_{2,2}+\dots+a_{n,n}=\operatorname{Tr}A$.

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@Ioannis Sorry, but there's no more elementary proof than the one I and Ted provided. And I don't think I'm able to divide it into more elementary steps than I did here. Therefore I think that you can't be helped. – yo' Oct 31 '13 at 0:05
Btw, this is not more advanced than Ted's proof. This is exactly the same, just each of the steps is written out. – yo' Oct 31 '13 at 0:11
There is a very simple proof for diagonalizable matrices that utlises the properties of the determinants and the traces. I am more interested in understanding your proofs though and that's what I have been striving to do. – JohnK Oct 31 '13 at 0:14

Let $A$ be a matrix. It has a Jordan Canonical Form, i.e. there is matrix $P$ such that $PAP^{-1}$ is in Jordan form. Among other things, Jordan form is upper triangular, hence it has its eigenvalues on its diagonal. It is therefore clear for a matrix in Jordan form that its trace equals the sum of its eigenvalues. All that remains is to prove that if $B,C$ are similar then they have the same eigenvalues.

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Trace is preserved under similarity and every matrix is similar to a Jordan block matrix. Since the Jordan block matrix has its eigenvalues on the diagonal, its trace is the sum (with multiplicity) of its eigenvalues.

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Every matrix $A$ can be written in the form $A=S^{-1}DS$ where $D$ is the diagonal matrix (with entries being eigenvalues) and $S$ is the matrix with columns being the associated eigenvectors.

Taking the trace on both sides we have:

$$\operatorname{tr}(A)=\operatorname{tr}(S^{-1}DS)=\operatorname{tr}(DSS^{-1})=\operatorname{tr}(D)=\sum \text{eigenvalues}.$$

Here I used the fact that $\operatorname{tr}(AB)=\operatorname{tr}(BA)$. As a plus we also have:

$$\det(A)=\det(S^{-1}DS)=\det(S^{-1})\det(D)\det(S)=\det(S)^{-1}\det(D)\det(S)=\det(D)=\prod \text{eigenvalues}$$

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What if the matrix is not diagonalizable, that is Jordan Normal Form? – Amzoti Mar 13 '14 at 15:56
This is wrong in so many ways (what if A is a matrix over the reals with complex eigenvalues?) and should be deleted or at least fixed. – roman Jun 30 '15 at 6:43

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