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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 $\psi: \bigwedge^n V \to ... 9 These answers require way too much machinery. By definition, the characteristic polynomial of an$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, ... 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 $$... 8 Both sides are continuous. A standard proof goes by showing this for diagonalizable matrices, and then using their density in M_n(\mathbb{C}). But actually, it suffices to triangularize$$ A=P^{-1}TP $$with P invertible and T upper-triangular. This is possible as soon as the characteristic polynomial splits, which is obviously the case in ... 7 Let A be the positive definite square root of X and B the positive definite square root of Y. You have$$ \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 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})\\ ... 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}$$then \sum\limits_{i = 1}^{n} a_{i, i} = 0. Hence, it doesn't matter what the elements which are not diagonal entries are, so each of those choices is free. Furthermore, we can select the ... 6 Let S_1, S_2 be two positive definite matrices. Let \Delta = S_2 - S_1 and for t \in [0,1], let$$\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 Yes, it is. ConsiderS(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 $$\left. \dfrac{d^2}{\partial t^2} ... 5 You don't need to choose any particular basis; all you need is the fact that \wedge V is graded and L_b raises the degree by 1, so if you choose any basis that respects the gradation, then L_b sends any basis vector to a different subspace, so the diagonal elements in such a basis all vanish. 5 As mentioned in the comments, the assertion "\operatorname{tr}(e^{tA}) = e^{\operatorname{tr}(tA)}" is simply false. On the other hand, the integration problem is straightforward. We have \exp(tA) = \sum_{n\geq 0} A^n \frac{t^n}{n!} for any finite-dimensional matrix A, since \exp has infinite radius of convergence. By linearity of integration, and ... 5 Let f(t)= \det(e^{tA}). Then f'(t)=D \det(e^{tA}) \cdot Ae^{tA}=\text{tr} \left(^t \text{com}(e^{tA})Ae^{tA} \right). But A and e^{tA} commute, and ^t\text{com}(e^{tA})e^{tA}=\det(e^{tA}) \operatorname{I}_n. Therefore, f'(t)=\text{tr}(A)f(t) and f(0)=1, hence f(t)=e^{\text{tr}(A)t}. For t=1, \det(e^{A})= e^{\text{tr}(A)}. 5 This is false. By the (finite-dimensional) spectral theorem, every symmetric matrix is diagonalizable, so we can write S=QDQ^{-1}. Thus we have$$\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 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, ... 4 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*} ... 4 In addition to the variety of useful perspectives already given: much as in Martin Brandenberg's answer, but less abstractly, while still coordinate-free... the map V\otimes V^*\rightarrow \mathrm{End}(V) induced from the bilinear map v\times \lambda\rightarrow (w\rightarrow \lambda(w)\cdot v) is a surjection for finite-dimensional vector spaces V. ... 4 By (AA^T)_{ii} you seem to mean the i-th term on the diagonal of AA^T. But instead what you've writen is already \mathrm{tr}(AA^T). Which is ok. Now, you just have to realize that both sums are the same, up to the order of the addends -which doesn't matter. For instance: you have a_{11}^2 on both sums, haven't you? Also a_{1m}^2 appears on both ... 4 To give you an idea of how to properly write these sort of proofs down, here's the proof. For a matrix X, let [X]_{ij} denote the (i,j) entry of X. Let A be m\times n and B be n\times m. Then \begin{align*} \mathrm{tr}\,(AB) &= \sum_{i=1}^n[AB]_{ii} \\ &= \sum_{i=1}^n\sum_{k=1}^m[A]_{ik}\cdot[B]_{ki} \\ &= ... 4 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 ... 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 ... 4 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 ... 4 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 ... 4 No. LetA = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\!, \quad B = \begin{bmatrix} 1 \\ & 0 \end{bmatrix}\!, \quad D = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\!, \quad c = 1.$$Then$$(A + B + cI)^{-1} A B D = \frac{1}{5}\begin{bmatrix} 2 & -1 \\ 4 & -2 \end{bmatrix}\!,$so$\mathop{\rm tr}((A + B + cI)^{-1} A B D) = ...

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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 ...

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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 ...

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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...) ...

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