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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 ...
$\newcommand{\tr}{\operatorname{tr}}$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 $\psi: \bigwedge^n V \to ... 12 These answers require way too much machinery. By definition, the characteristic polynomial of an$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, ... 10 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 ... 8 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 ... 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 $$... 8 If the matrices are non-singular, then writing A=-BAB^{-1} and taking the trace, we get \mathrm{tr}A=-\mathrm{tr}A. Hence \mathrm{tr}A=0, and the procedure for B is analogous. Next compute the determinant of both sides of AB=-BA: this yields \mathrm{det}\,A\,\mathrm{det}\,B=(-1)^N\mathrm{det}\,B\,\mathrm{det}\,A, where N stands for size of ... 7 Let A be the positive definite square root of X and B the positive definite square root of Y. You have$$ \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 ... 7 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})\\ ... 7 Yes, it is. Consider S(t) = A + t B where A is symmetric positive definite and B is 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} ... 6 LetS_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 Recall that there are n^2 total entries in the matrix. The matrix (which we call A) is subject to the condition that ifA = [a_{i, j}]_{1 \leq i, j \leq n}$$then \sum\limits_{i = 1}^{n} a_{i, i} = 0. Hence, it doesn't matter what the elements which are not diagonal entries are, so each of those choices is free. Furthermore, we can select the ... 6 What is true is that the expansion of the characteristic polynomials is given by traces of powers of the matrix A; explicitly, the characteristic polynomial \chi_A(T)=\det(T\cdot{\rm Id}-A) is given by$$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 $$... 5 This is false. By the (finite-dimensional) spectral theorem, every symmetric matrix is diagonalizable, so we can write S=QDQ^{-1}. Thus we have$$\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 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 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, ... 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 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 ... 5 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 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. 4 There's no failure, unless you count assuming that such a matrix X exists for any A. You've in fact given a proof that if \det{A}\leq0, there is no X (with real trace) such that e^X=A. If you allow \operatorname{Tr}(X) to be complex (so it is possible for e^{\operatorname{Tr}(X)} to be negative), then there does exist X with e^X=A for any ... 4 Hint: the trace is multiplicative, i.e. \mbox{Tr}(AB)=\mbox{Tr}(BA), as soon as the scalar field (or ring) is commutative. In particular, for every invertible matrix P and every matrix A in M_n(\mathbb{R}) or M_n(\mathbb{C}), we have \mbox{Tr}\;(PAP^{-1})=\mbox{Tr}\; A. $$4 For (1), see the citation in my answer to a previous question. In particular, yes, the set of all traceless matrices are precisely the set of all commutators, regardless of the underlying field. The exercise in Hoffman and Kunze asks whether the subspace of all traceless matrices is equal to the subspace spanned by all commutators. This is different from ... 4 At least in the best circumstances, it is easy to prove that the trace/restriction map loses {\ell\over p}+\epsilon in "Sobolev units" (for arbitrarily small \epsilon>0) where codimension is \ell. And an easy extension of Sobolev imbedding theorems shows that (for example, for L^2 Sobolev spaces so I don't mess up the indexing shift...) ... 4 Let A be symmetric positive definite matrix hence \exists a diagonal matrix D whose diagonal entries are nonzero and A=P D P^{-1} so A^{-1} = P D^{-1} P^{-1} and Tr(A^{-1})= Tr(D^{-1}). Now D being diagonal matrix with non zero diagonal entries D^{-1} has diagonal entries reciprocal of the diagonal entries of D so Tr(D^{-1}) is sum of ... 4 For clarification, following the question, let us emphasize some notations first. Let M_n be the collection of n\times n complex valued matrices, let H_n be the collection of n\times n complex valued Hermitian matrices, and given X,Y\in H_n, denote X\ge 0\Leftrightarrow X is positive semidefinite, and X\ge Y\Leftrightarrow X-Y\ge 0. Given ... 4 If A=(a_{ij}) and B=(b_{ij}) and C=B^TA=(c_{ij}) then$$(c)_{ij}=\sum_{k=1}^m b_{ki}a_{kj}\mathrm{tr}(B^TA)=\sum_{i=1}^n c_{ii}=\sum_{i=1}^n\sum_{k=1}^m b_{ki}a_{ki}$so we see that$\langle.,.\rangle$is an inner product (Euclidian) by identifying$\mathcal M_{m\times n}(\mathbb R)$to$\mathbb R^{m\times n}$. 4 You may do it by first computing matrix powers and then you may calculate whatever you want. Now question is how to calculate matrix power for a given matrix, say$A$? Your goal here is to develop a useful factorization$A = PDP^{-1}$, when$A$is$n\times n$matrix.The matrix$D\$ is a diagonal matrix (i.e. entries off the main diagonal are all zeros). Then ...