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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2This has a detailed proof: https://www.adelaide.edu.au/mathslearning/play/seminars/evaluemagictrickshandout.pdf – Anurag May 15 '19 at 14:57

2Is it true for non algebraically closed field? – user595419 Mar 07 '20 at 03:49

2This is what I found online: Hopefully, it will help http://www.math.harvard.edu/~knill/teaching/math19b_2011/handouts/lecture28.pdf – Ramuel May 01 '19 at 19:07
8 Answers
These answers require way too much machinery. By definition, the characteristic polynomial of an $n\times n$ matrix $A$ is given by $$p(t) = \det(AtI) = (1)^n \big(t^n  (\text{tr} A) \,t^{n1} + \dots + (1)^n \det A\big)\,.$$ 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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49The definition of the characteristic polynomial I learned is just only $\det{(AtI)}$. Why the coefficient of the $t^{n1}$ term is $\text{tr }A$? – bfhaha Apr 08 '18 at 22:08

29@bfhaha The coefficient of $t^{n1}$ must come from $(a_{11}t)\cdots (a_{nn}t)$, because if we expand $A\lambda I$ along a row $i$, we see that any term involving an offdiagonal element $[At I]_{ij}$ eliminates $a_{ii}t$ and $a_{jj}t$, and hence any such term does not involve $t^{n1}$. Therefore the coefficient of $t^{n1}$ must come from the product of the diagonal elements, and so the coefficient is $(1)^{n1}(a_{11}+\cdots+a_{nn})$. – Elias Aug 14 '18 at 07:34

Hello! My understanding of the characteristic polynomial is from the determinant itself. So how did one figure out this will be the same the second form i.e $p(t) = (t\lambda_1)(t\lambda_2)...$ – DuttaA Dec 30 '20 at 17:00

3@DuttaA Because the eigenvalues $\lambda_i$ are by definition the roots of this polynomial. – Ted Shifrin Dec 30 '20 at 17:18

@Elias what does it mean to "expand along a row $i$"? I'm still not clear why the coefficient of $t^{n1}$ is equal to the trace of $A$. – Max Sep 19 '21 at 08:32

@Max They mean to rewrite $\det(AtI)$ in terms of the cofactors along a row of $AtI$. This is called Laplace expansion of the determinant. – Michael L. May 03 '22 at 19:04
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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25I would try to elaborate a bit. For every matrix $A$ there exists a non singular matrix $P$ such that $PAP^{1} = J$ where $J$ has [Jordan canonical form](https://en.wikipedia.org/wiki/Jordan_normal_form). Now using $tr(ABC) = tr(CAB) = tr(BCA)$ (which is true whenever the products are defined), we obtain $tr(A) = tr(P^{1}JP) = tr(PP^{1}J) = tr(J) = \sum_i \lambda_i$ where $\lambda_i$ are the eigenvalues of $A$. – them Aug 14 '16 at 10:08

1Very elegant :) Also such tool can be used to show that det(A) ofr any matrix A is the product of eigenvalues det(A). – Konstantin Burlachenko Jan 24 '18 at 13:32

@bruziuz can you please tell me how can I show that determinant of a matrix in Jordan form is product of its diagonal entries? – Abhay Oct 21 '19 at 17:00

1@Abhay, you are posting a comment to an answer from 6 years ago. Fortunately, I am still around to answer it. The answer is: Jordan form is, among other things, upper triangular. All upper triangular matrices have their determinant as the product of the diagonal entries. This can be proved by recursively [Laplace expanding](https://en.wikipedia.org/wiki/Determinant#Laplace's_formula_and_the_adjugate_matrix) on the first column. – vadim123 Oct 21 '19 at 17:08

@vadim123 thank you, your answer to above post really helped me. Thanks again for clarifying the determinant bit. – Abhay Oct 21 '19 at 17:13

@Abhay det(P^1 J P)=det(P^1)det(J)det(P)=det(J) J is block diagonal and it's blocks has bidiagonal form, but at least J is upper triangular. In particular, for any triangular matrix determinant is a product of all entries in diagonal. – Konstantin Burlachenko Jan 10 '20 at 08:50
I'll try to show it another way. We know that if we have a polynomial $x^n+b_{n1} x^{n1} + \dots +b_1 x+ b_0$, then $(1)^{n1} b_{n1}$ is the sum of the roots of this polynomial. (Socalled Vieta's formulas) In our case, the polynomial is $\det(tIA)$ and we have $(1)^{n1} b_{n1}=\lambda_1+\lambda_2+\dots+\lambda_n$.
$\def\S{\mathcal{S}_n}$ 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 $M=tIA$. To get the power $t^{n1}$ for a permutation, we need this permutation to choose at least $n1$ diagonal elements, i.e., to have $\pi(i)=i$ for at least $n1$ values of $i$. However, once you know the value of a permuation on $n1$ inputs, you know the last one as well. This means, that to get the coefficient of $t^{n1}$, we need to consider only the identity permutation.
So far we got that $b_{n1}$ is the coefficient of $t^{n1}$ in $(ta_{1,1})(ta_{2,2})\dots(ta_{n,n})$ (this is the term of the sum above corresponding to the identity permutation). Therefore $(1)^{n1}b_{n1} = a_{1,1}+a_{2,2}+\dots+a_{n,n}=\operatorname{Tr}A$.
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1@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 00:05

2Btw, 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 00:11

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 00:14
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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Here is another proof. First of all, by definition, we have that the characteristic polynomial of the $n\times n$ matrix $A=[a_{ij}]$ is given by $P_A(x)=\det(xI_nA)$. Let $P_A(x)=x^nb_1x^{n1}+b_2x^{n2}\dots$. By Viete's formula the sum of eigenvalues is $b_1$. We have to prove that $b_1=\hbox{trace}(A)$. Substituting $x$ with $\frac{1}{x}$ for every real nonzero $x$ we get $$\det\left(\frac{1}{x}\left(I_nxA\right)\right)=\frac{1b_1x+b_2x^{2}\dots}{x^n},$$ or equivalently $$\det(I_nxA)=1b_1x+b_2x^2\dots$$ for any nonzero real $x$. But then the left side and right side polynomials from the above equation coincide for all real $x$. A short explanation: if for two polynomials $f$ and $g$ we have $f(x)=g(x)$ for any nonzero $x$, then the polynomial $h(x):=f(x)g(x)$ has an infinity of zeroes, thus being the identically zero polynomial; it follows that $f(x)=g(x)$ for all $x$. Let's denote now $$f(x):=\det(I_nxA)$$ and $$g(x):=1b_1x+b_2x^2\dots.$$ We have seen that $f$ and $g$ are equal functions (polynomials). We then have $f'(x)=g'(x)$ for all $x$. Obviously, $g'(x)=b_1+2b_2x\dots$, therefore $g'(0)=b_1$. On the other side, from $$f(x)=\left\begin{array}{cccc} 1a_{11}x&a_{12}x&\dots&a_{1n}x\\ a_{21}x&1a_{22}x&\dots&a_{2n}x\\ \dots&\dots&\dots&\dots\\ a_{n1}x&a_{n2}x&\dots&1a_{nn}x\end{array}\right$$ by the rule of differentiating determinants we get $$f'(x)=\left\begin{array}{cccc} a_{11}&a_{12}&\dots&a_{1n}\\ a_{21}x&1a_{22}x&\dots&a_{2n}x\\ \dots&\dots&\dots&\dots\\ a_{n1}x&a_{n2}x&\dots&1a_{nn}x\end{array}\right+\dots+\left\begin{array}{cccc} 1a_{11}x&a_{12}x&\dots&a_{1n}x\\ a_{21}x&1a_{22}x&\dots&a_{2n}x\\ \dots&\dots&\dots&\dots\\ a_{n1}&a_{n2}&\dots&a_{nn}\end{array}\right.$$ It follows that $$f'(0)=\left\begin{array}{cccc} a_{11}&a_{12}&\dots&a_{1n}\\ 0&1&\dots&0\\ \dots&\dots&\dots&\dots\\ 0&0&\dots&1\end{array}\right+\dots+\left\begin{array}{cccc} 1&0&\dots&0\\ 0&1&\dots&0\\ \dots&\dots&\dots&\dots\\ a_{n1}&a_{n2}&\dots&a_{nn}\end{array}\right=$$ $$=(a_{11}+\dots+a_{nn})=\hbox{trace}(A).$$ Since we have $f'(0)=g'(0)$ we get $b_1=\hbox{trace}(A)$.
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Let $A \in M_{n}(\mathbb{C})$. Then $\text{tr}(A) = \text{tr}(PTP^{1})$ where $T$ is an upper triangular matrix and $P$ is invertible$^{1}$. Thus $\text{tr}(A) = \text{tr}(PTP^{1}) {=} \text{tr}(P^{1}PT) = \text{tr}(T)$. The result follows since the diagonal entries of $T$ are the eigenvalues of $A$.
$^{1}$ The existence of matrices $T$ and $P$ follows from the fact that $\mathbb{C}$ is algebraically closed!
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By the Schur decomposition, any matrix $A$ is unitarily similar to an upper triangular matrix $T$. Being similar, $A$ and $T$ have the same trace and the same eigenvalues. Moreover, the diagonal entries of $T$ are equal to its eigenvalues (since $T$ is triangular). The stated result follows by calculating the trace of $T$. See https://www.statlect.com/matrixalgebra/propertiesofeigenvaluesandeigenvectors.
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Let $\mathbf{A}$ be a $k \times k$ symmetric matrix and $\mathbf{x}$ be a $k \times 1$ vector. Then
(a) $\mathbf{x'Ax}$ = tr($\mathbf{x'Ax}$) = tr($\mathbf{Axx'}$)
(b) tr($\mathbf{A}$) = $\Sigma_{i=1}^k \lambda_i$, where the $\lambda_i$ are the eigenvalues of $\mathbf{A}$.
For Part a, we note that $\mathbf{x'Ax}$ is a scalar, so $\mathbf{x'Ax}$ = tr($\mathbf{x'Ax}$). We know that tr($\mathbf{BC}$) = tr($\mathbf{CB}$) for any two matrics $\mathbf{B}$ and $\mathbf{C}$ of dimensions $m \times k$ and $k \times m$, respectively. This follows because $\mathbf{BC}$ has $\Sigma_{j=1}^k b_{ij}c_{ji}$ as its ith diagonal element, so tr($\mathbf{BC}$) = $\Sigma_{i=1}^m ( \Sigma_{j=1}^k b_{ij}c_{ji})$. Similarly, the jth diagonal element of $\mathbf{CB}$ is $\Sigma_{i=1}^m c_{ji}b_{ij}$, so tr($\mathbf{CB}$) = $\Sigma_{j=1}^k ( \Sigma_{i=1}^m c_{ji}b_{ij})$ = $\Sigma_{i=1}^m ( \Sigma_{j=1}^k b_{ij}c_{ji})$ = tr($\mathbf{BC}$).
Let $\mathbf{x'}$ be the matrix $\mathbf{B}$ with m = 1, and let $\mathbf{Ax}$ play the role of the matrix $\mathbf{C}$. Then tr($\mathbf{x'(Ax)}$) = tr($\mathbf{(Ax)x'}$), and the result follows.
Part b is proved by using the spectral decomposition to write $\mathbf{A=P' \Lambda P}$, where $\mathbf{PP'=I}$ and $\mathbf{\Lambda}$ is a diagonal matrix with entries $\lambda_1$,$\lambda_2$,...,$\lambda_k$. Therefore, tr($\mathbf{A}$) = tr($\mathbf{P' \Lambda P}$) = tr($\mathbf{\Lambda P P'}$) = tr($\mathbf{\Lambda}$) = $\lambda_1 + \lambda_2 + \lambda_k$.
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