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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, ...

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The trace of an endomorphism $f : X \to X$ of a dualizable object $X$ in a monoidal category is the composition $1 \xrightarrow{\eta} X \otimes X^* \xrightarrow{f \otimes \mathrm{id}} X \otimes X^* \cong X^* \otimes X \xrightarrow{\epsilon} 1$. This coincides with the usual definition in the category of vector spaces. There is a more general categorical ...

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$\newcommand{\tr}{\operatorname{tr}}$Here is an exterior algebra approach. Let $V$ be an $n$-dimensional vector space and let $\tau$ be a linear operator on $V$. The alternating multilinear map $$(v_1,\dots,v_n) \mapsto \sum_{k=1}^n v_1 \wedge\cdots\wedge \tau v_k \wedge\cdots\wedge v_n$$ induces a unique linear operator $\psi: \bigwedge^n V \to ... 11 Both sides are continuous. A standard proof goes by showing this for diagonalizable matrices, and then using their density in$M_n(\mathbb{C})$. But actually, it suffices to triangularize $$A=P^{-1}TP$$ with$P$invertible and$T$upper-triangular. This is possible as soon as the characteristic polynomial splits, which is obviously the case in ... 10 If the matrices are non-singular, then writing$A=-BAB^{-1}$and taking the trace, we get$\mathrm{tr}A=-\mathrm{tr}A$. Hence$\mathrm{tr}A=0$, and the procedure for$B$is analogous. Next compute the determinant of both sides of$AB=-BA$: this yields$\mathrm{det}\,A\,\mathrm{det}\,B=(-1)^N\mathrm{det}\,B\,\mathrm{det}\,A$, where$N$stands for size of ... 10 This statement is false : take$S = I$for example. It would give us$2^n = 1 + n$. 9 Yes, it is. Consider$S(t) = A + t B$where$A$is symmetric positive definite and$Bis symmetric. It is enough to show that $$\left.\dfrac{d^2}{d t^2} \text{Tr}(S(t)^{-1})\right|_{t=0} \ge 0$$ Now $$S(t)^{-1} = (A (I + t A^{-1} B))^{-1} = A^{-1} - t A^{-1} B A^{-1} + t^2 A^{-1} B A^{-1} B A^{-1} + \ldots$$ so \left. \dfrac{d^2}{\partial t^2} ... 9 The proof in Martin Brandenburg's answer may look scary but it is secretly about moving beads around on a string. You can see all of the relevant pictures in this blog post and in this blog post. The proof using pictures is the following: In the first step g gets slid down on the right and in the second step g gets slid up on the left. You can also ... 9 Hint: Use that every complex matrix has a jordan normal form and that the determinant of a triangular matrix is the product of the diagonal. use that \exp(A)=\exp(S^{-1} J S ) = S^{-1} \exp(J) S And that the trace doesn't change under transformations. \begin{align*} \det(\exp(A))&=\det(\exp(S J S^{-1}))\\ &=\det(S \exp(J) S^{-1})\\ ... 8 Hint Compare the characteristic polynomials of AB and BA. The determinant (whence characteristic polynomials) admits basis-free definitions. We have \left(\matrix{I&A\\B&tI}\right)\left(\matrix{tI&-A\\0&I}\right)=\left(\matrix{tI&0\\*&tI-BA}\right) $$and$$ ... 8 LetA$be a matrix. It has a Jordan Canonical Form, i.e. there is matrix$P$such that$PAP^{-1}$is in Jordan form. Among other things, Jordan form is upper triangular, hence it has its eigenvalues on its diagonal. It is therefore clear for a matrix in Jordan form that its trace equals the sum of its eigenvalues. All that remains is to prove that if ... 7 Let$A$be the positive definite square root of$X$and$B$the positive definite square root of$Y$. You have $$\mbox{tr}(XY)=\mbox{tr}(AABB)=\mbox{tr}(BAAB)=\mbox{tr}((AB)^*AB)>0.$$ Indeed, the latter is the sum of all$c_{i,j}^2$where$c_{i,j}=(AB)_{i,j}$. So it is nonnegative. And if it were zero, this would imply$AB=0$hence$A=B=0$since ... 6 What is true is that the expansion of the characteristic polynomials is given by traces of powers of the matrix$A$; explicitly, the characteristic polynomial$\chi_A(T)=\det(T\cdot{\rm Id}-A)$is given by $$T^n-{\rm tr}(A)T^{n-1}+\frac{{\rm tr}(A)^2-{\rm tr}(A^2)}{2}T^{n-2}-\frac{{\rm tr}(A)^3-3{\rm tr}(A){\rm tr}(A^2)+2{\rm tr}(A^3)}{6}T^{n-3}+\cdots$$ ... 6 Recall that there are$n^2$total entries in the matrix. The matrix (which we call$A$) is subject to the condition that if $$A = [a_{i, j}]_{1 \leq i, j \leq n}$$ then$\sum\limits_{i = 1}^{n} a_{i, i} = 0$. Hence, it doesn't matter what the elements which are not diagonal entries are, so each of those choices is free. Furthermore, we can select the ... 6 As mentioned in the comments, the assertion "$\operatorname{tr}(e^{tA}) = e^{\operatorname{tr}(tA)}$" is simply false. On the other hand, the integration problem is straightforward. We have$\exp(tA) = \sum_{n\geq 0} A^n \frac{t^n}{n!}$for any finite-dimensional matrix$A$, since$\exp$has infinite radius of convergence. By linearity of integration, and ... 6 I assume you want the trace of a matrix$A\in M_n(F)$to be defined as the sum of the diagonal elements and that you take the coefficients in a (commutative) field$F$. Here is an approach using only basic facts about bases and matrices. 1) Recall the trace is commutative$\mathrm{tr}(AB)=\mathrm{tr}(BA)$, as shown by the usual computations. In particular ... 6 Note that, when$D$is diagonal: $$(DA)_{ii} = D_{ii} A_{ii}$$ So$tr(DA) = \sum_{i=1}^n D_{ii} A_{ii}$. About the best bound you can do for this is the Cauchy-Schwarz inequality, i.e. $$|tr(DA)| \leq \left ( \sum_{i=1}^n D_{ii}^2 \right )^{1/2} \left ( \sum_{i=1}^n A_{ii}^2 \right )^{1/2}$$ If you want a result in terms of traces, you can use the fact ... 6 Let$A$be symmetric positive definite matrix hence$\exists$a diagonal matrix$D$whose diagonal entries are nonzero and$A=P D P^{-1}$so$A^{-1} = P D^{-1} P^{-1}$and$Tr(A^{-1})= Tr(D^{-1})$. Now$D$being diagonal matrix with non zero diagonal entries$D^{-1}$has diagonal entries reciprocal of the diagonal entries of$D$so$Tr(D^{-1})$is sum of ... 6 Let$S_1$,$S_2$be two positive definite matrices. Let$\Delta = S_2 - S_1$and for$t \in [0,1], let $$\phi = (S_1 + t\Delta)^{-1} = ((1 - t) S_1 + t S_2)^{-1}$$ We have: \begin{align} & \frac{d}{dt} \phi \;= - \phi \Delta \phi\\ \implies & \frac{d}{dt} \phi^2 \;= - \phi \Delta \phi^2 - \phi^2 \Delta \phi\\ \implies & ... 6 The reason that the equality Tu=u_{|\partial \Omega} is stated for functions in C(\overline{\Omega}) is that for other functions it is not clear what u_{|\partial \Omega} means. Of course, we can take any function u\in W^{1,p}(\Omega) and extend its domain to \overline{\Omega} by letting u be equal to Tu on the boundary. This will be an ... 6 The following is a simple combinatorial interpretation of this identity. Not exactly what you asked for, but still fun and relevant. Suppose we have two sets S,T with functions g: S \to T and f : T \to S. Then f\circ g : S \to S and g\circ f: T \to T are endo-functions of S and T respectively. Now consider \text{Fix}(f\circ g) \subseteq S, ... 6 Not true. Try A = B = \pmatrix{1 & t\cr t & 1\cr} for 0 < t < 1. 6 Let f(t)= \det(e^{tA}). Then f'(t)=D \det(e^{tA}) \cdot Ae^{tA}=\text{tr} \left(^t \text{com}(e^{tA})Ae^{tA} \right). But A and e^{tA} commute, and ^t\text{com}(e^{tA})e^{tA}=\det(e^{tA}) \operatorname{I}_n. Therefore, f'(t)=\text{tr}(A)f(t) and f(0)=1, hence f(t)=e^{\text{tr}(A)t}. For t=1, \det(e^{A})= e^{\text{tr}(A)}. 5 Yes, it holds true. Let A be a n\times m and B be a m \times n matrix over the commutative ring R, we have \begin{align*} \mathrm{tr}(AB) &= \sum_{i=1}^n (AB)_{ii}\\ &=\sum_{i=1}^n \sum_{j=1}^m A_{ij}B_{ji}\\ &= \sum_{j=1}^m \sum_{i=1}^n B_{ji}A_{ij}\\ &= \sum_{j=1}^m (BA)_{jj}\\ &= \mathrm{tr}(BA) \end{align*} ... 5 X^\top=-SXS^{-1}, so \operatorname{tr}X=\operatorname{tr}X^\top = -\operatorname{tr}(SXS^{-1})=-\operatorname{tr} X. 5 You don't need to choose any particular basis; all you need is the fact that \wedge V is graded and L_b raises the degree by 1, so if you choose any basis that respects the gradation, then L_b sends any basis vector to a different subspace, so the diagonal elements in such a basis all vanish. 5 The trace of a matrix A\in\mathbb R^{n\times n} is nothing but the sum of its eigenvalues. So Ker(Trace) is the set of matrices with vanishing sum of eigenvalues, and its dimension is n^2-1. The quotient space {\mathcal M}_n/Ker(Trace) is a subspace of dimension 1 isomorphic to \mathbb R. The elements of this quotient space are classes of ... 5 The trace of a matrix M is 0 if and only if the sum of the elements on the (main) diagonal of M is 0. Since the dimension of all n\times n matrices is n^2 and the dimension of its image \mathbb R is 1 (see below), we know that the dimension of the kernel of \mathrm{tr} is n^2-1. (That follows from the fact that the dimension of the image ... 5 This is false. By the (finite-dimensional) spectral theorem, every symmetric matrix is diagonalizable, so we can write S=QDQ^{-1}. Thus we have\det(I+S)=\det(Q^{-1}(I+S)Q)=\det(I+D)=\prod\limits_{i=1}^n (1+d_i)\\\ne 1+\sum\limits_{i=1}^n d_i=1+\mathrm{trace}(D)=1+\mathrm{trace}(S).$Perhaps the formula$\prod\limits_{i=1}^n (1+d_i)$may be of use, ... 5 I'll try to show it another way. We know that if we have a polynomial$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 ...

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