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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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8 Answers 8

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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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    $\begingroup$ The definition of the characteristic polynomial I learned is just only $\det{(A-tI)}$. Why the coefficient of the $t^{n-1}$ term is $-\text{tr }A$? $\endgroup$
    – bfhaha
    Apr 8, 2018 at 22:08
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    $\begingroup$ @bfhaha The coefficient of $t^{n-1}$ 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 off-diagonal element $[A-t I]_{ij}$ eliminates $a_{ii}-t$ and $a_{jj}-t$, and hence any such term does not involve $t^{n-1}$. Therefore the coefficient of $t^{n-1}$ must come from the product of the diagonal elements, and so the coefficient is $(-1)^{n-1}(a_{11}+\cdots+a_{nn})$. $\endgroup$
    – Elias
    Aug 14, 2018 at 7:34
  • $\begingroup$ 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 $p(t) = (t-\lambda_1)(t-\lambda_2)...$ $\endgroup$
    – DuttaA
    Dec 30, 2020 at 17:00
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    $\begingroup$ @DuttaA Because the eigenvalues $\lambda_i$ are by definition the roots of this polynomial. $\endgroup$ Dec 30, 2020 at 17:18
  • $\begingroup$ @Elias what does it mean to "expand along a row $i$"? I'm still not clear why the coefficient of $t^{n-1}$ is equal to the trace of $A$. $\endgroup$
    – Max
    Sep 19, 2021 at 8:32
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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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    $\begingroup$ I 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. 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 $A$. $\endgroup$
    – them
    Aug 14, 2016 at 10:08
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    $\begingroup$ Very elegant :) Also such tool can be used to show that det(A) ofr any matrix A is the product of eigenvalues det(A). $\endgroup$ Jan 24, 2018 at 13:32
  • $\begingroup$ @bruziuz can you please tell me how can I show that determinant of a matrix in Jordan form is product of its diagonal entries? $\endgroup$
    – Abhay
    Oct 21, 2019 at 17:00
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    $\begingroup$ @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 on the first column. $\endgroup$
    – vadim123
    Oct 21, 2019 at 17:08
  • $\begingroup$ @vadim123 thank you, your answer to above post really helped me. Thanks again for clarifying the determinant bit. $\endgroup$
    – Abhay
    Oct 21, 2019 at 17:13
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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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    $\begingroup$ @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. $\endgroup$
    – yo'
    Oct 31, 2013 at 0:05
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    $\begingroup$ Btw, this is not more advanced than Ted's proof. This is exactly the same, just each of the steps is written out. $\endgroup$
    – yo'
    Oct 31, 2013 at 0:11
  • $\begingroup$ 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. $\endgroup$
    – JohnK
    Oct 31, 2013 at 0:14
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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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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 $P_A(x)=\det(xI_n-A)$. Let $P_A(x)=x^n-b_1x^{n-1}+b_2x^{n-2}-\dots$. 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 non-zero $x$ we get $$\det\left(\frac{1}{x}\left(I_n-xA\right)\right)=\frac{1-b_1x+b_2x^{2}-\dots}{x^n},$$ or equivalently $$\det(I_n-xA)=1-b_1x+b_2x^2-\dots$$ for any non-zero 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 non-zero $x$, 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 $$f(x):=\det(I_n-xA)$$ and $$g(x):=1-b_1x+b_2x^2-\dots.$$ 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 $$f(x)=\left|\begin{array}{cccc} 1-a_{11}x&-a_{12}x&\dots&-a_{1n}x\\ -a_{21}x&1-a_{22}x&\dots&-a_{2n}x\\ \dots&\dots&\dots&\dots\\ -a_{n1}x&-a_{n2}x&\dots&1-a_{nn}x\end{array}\right|$$ by the rule of differentiating determinants we get $$f'(x)=\left|\begin{array}{cccc} -a_{11}&-a_{12}&\dots&-a_{1n}\\ -a_{21}x&1-a_{22}x&\dots&-a_{2n}x\\ \dots&\dots&\dots&\dots\\ -a_{n1}x&-a_{n2}x&\dots&1-a_{nn}x\end{array}\right|+\dots+\left|\begin{array}{cccc} 1-a_{11}x&-a_{12}x&\dots&-a_{1n}x\\ -a_{21}x&1-a_{22}x&\dots&-a_{2n}x\\ \dots&\dots&\dots&\dots\\ -a_{n1}&-a_{n2}&\dots&-a_{nn}\end{array}\right|.$$ 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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    $\begingroup$ Great first answer! Good job! $\endgroup$
    – user370967
    Jun 7, 2017 at 19:51
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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/matrix-algebra/properties-of-eigenvalues-and-eigenvectors.

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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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