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13

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 ...


12

$\newcommand{\tr}{\operatorname{tr}}$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

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} ...


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})\\ ...


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

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 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

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 & ...


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

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

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

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

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

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 $\text{Fix}(f\circ g) \subseteq S$, ...


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

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

Let $A=\sqrt{S}$, and equip the space of $n\times n$ real matrices with the usual Euclidean scalar product. Then $$\hbox{Tr}(RZ)= \hbox{Tr}(RUA^2V^T)=\hbox{Tr}((RUA)(VA)^T)=\langle RUA,VA\rangle$$ By the Cauchy-Schwartz inequality, we get $$\hbox{Tr}(RZ)\leq \Vert RUA \Vert_2 \Vert VA \Vert_2= \Vert A \Vert_2 \Vert A \Vert_2 =\hbox{Tr}(AA^T)=\hbox{Tr}(S)$$ ...


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

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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