Proof that the trace of a matrix is the sum of its eigenvalues 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.
 A: 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.
A: 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 $(-1)^{n-1}b_{n-1} = a_{1,1}+a_{2,2}+\dots+a_{n,n}=\operatorname{Tr}A$.
A: 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 $\text{tr}A = \lambda_1 + \dots + \lambda_n$.
A: 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.
A: 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($\mathbf{\Lambda}$) = $\lambda_1 + \lambda_2 + \lambda_k$.
A: 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 $b_1=\hbox{trace}(A)$.
A: 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!
A: 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.
