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26 votes

Geometric interpretation of trace

People have almost said this but not quite: Take any linear transformation $A$ of a finite-dimensional real vector space $V$. Let each point $v$ in $V$ start moving at the velocity $Av$. Then the ...
18 votes

Matrix trace & norm

For hermitian $B$, the inequality is very easy to prove: use that $\|B\| - B$ is positive semidefinite together with the fact that the product of two positive semidefinite matrices has nonnegative ...
Tobias Fritz's user avatar
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17 votes

Matrix trace & norm

Expanding my comment into an answer, which offers a more general result. Theorem (von Neumann). Let $A$ and $B$ be arbitrary $n\times n$ complex matrices. Then, $$|\text{trace}(AB)| \le \sum_{i=1}^...
Suvrit's user avatar
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15 votes
Accepted

Trace of non-commutable matrices

Your conjecture is a special case of the following result which essentially follows from the Lieb-Thirring inequality. Let $A$ and $B$ be Hermitian matrices. Then, for every positive integer $p$ we ...
Suvrit's user avatar
  • 28.4k
12 votes
Accepted

Reconstruct a matrix from its traces

Unfortunately even the generic situation is bad. Since we know the eigenvalues, we should search for the orthonormal system of eigenvectors $v_i$ of $A$. We have ($e_i$ is the standard basis) $$ Tr(A^...
fedja's user avatar
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11 votes

Geometric interpretation of characteristic polynomial

I am reluctant to answer a question this old that already has a very nice answer, however, looking at the title the first thing that comes to my mind is something quite different from the existing ...
Sean Lawton's user avatar
  • 8,414
11 votes

Geometric interpretation of trace

I like the following perspective: Up to scalar, trace is the only linear operator $\text{M}(n,k) \stackrel{t}{\to} k $ such that $t(AB) = t(BA)$. If one likes vector field theory, this is the only ...
11 votes

Matrix trace & norm

A proof of $$\lambda_n(B)\,{\rm tr}\,A\leq {\rm tr}\,(AB)\leq\lambda_1(B)\,{\rm tr}\,A$$ where $\lambda_n$ is the $n$-th largest eigenvalue of $B$ so $||B||=\lambda_1(B)$ and $A$, $B$ are positive ...
Carlo Beenakker's user avatar
10 votes

Trace of non-commutable matrices

Not to take anything away from Suvrit's answer, but this is actually much simpler. First, we can assume $M_1$ is diagonal. Call it $diag(x_1, \dotsc, x_i).$ Then the difference between the LHS and the ...
Igor Rivin's user avatar
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10 votes
Accepted

If I multiply the coefficients of a trace-class operator with bounded complex numbers is it still trace class?

It's a little more complicated than I thought! Frederik Ravn Klausen pointed out an error. Still, I maintain that the product needn't even be bounded. As the answer to this question shows, in $M_n$ ...
Nik Weaver's user avatar
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9 votes

Matrix trace & norm

We can work in an orthonormal basis in which $A$ is diagonal. In that case the diagonal entries of $AB$ are $a_{ii}b_{ii}$, so $$ tr(AB) \leq|tr(AB)|=\left|\sum a_{ii}b_{ii}\right| \leq \sum a_{ii}|...
Mike Jury's user avatar
  • 2,351
8 votes

Geometric interpretation of trace

Was surprised not to see this here yet. Let $V$ be an $n$-dimensional real vector space with inner product. Any linear transformation $f:V \to V$ can be decomposed into $$ f = \left(\tfrac{\textrm{tr}...
8 votes

Geometric interpretation of trace

Taking a broad view of the question, here are some particular geometric interpretations of the trace with respect to certain domains: $\mathrm{SL}(2, \mathbb{R})$ acts by isometries on the upper half-...
8 votes
Accepted

Completely bounded norm for unital maps with completely positive sections

Unfortunately, the answer is no. Suppose $\Phi : M_4 \rightarrow M_2$ and $\Psi : M_2 \rightarrow M_4$ are given by $$\Phi\left(\left[\begin{array}{cc} A & B\\ C& D\end{array}\right]\right) =...
Chris Ramsey's user avatar
  • 3,954
8 votes
Accepted

Is there a converse to the Brauer–Nesbitt theorem?

Not always — e.g. $g(1)$ should be an integer. The desired description is given in Helling, H., Eine Kennzeichnung von Charakteren auf Gruppen und assoziativen Algebren, Commun. Algebra 1, 491-501 (...
Francois Ziegler's user avatar
8 votes

If I multiply the coefficients of a trace-class operator with bounded complex numbers is it still trace class?

Nik Weaver's answer gives a nice counter-example. Let me just say a few words of context. Kernels $K$ for which $KT$ is trace-class for all trace-class $T$ are called Schur multipliers. (Not to be ...
Matthew Daws's user avatar
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7 votes
Accepted

Bound on sum of $n$th super-diagonal entries in a $2n$ by $2n$ PSD matrix

If $\left(\begin{array}{} A & B \\ B^T & C \end{array}\right)\succeq 0$ then there exists a contraction $K$ (i.e. $\lambda_n(K)=\|K\| \leq 1$) such that $B = A^{1/2} K C^{1/2}$ (e.g. see ...
Mahdi's user avatar
  • 1,991
7 votes

Inequality for trace of a symmetric product?

If the $v_i$ are orthonormal, then yes. If the $v_i$ aren't assumed to be orthogonal, they can cluster around the dominant eigenvector and cause a counterexample. Let the eigenvalues of $A$ be $\{\...
MTyson's user avatar
  • 1,583
7 votes

Trace inequality under consideration of definiteness

Write $$G=\left( \begin{array}{ccc} a & b & c \\ b & d & e \\ c & e & f \\ \end{array} \right) $$ and, without loss of generality, $$ U=\left( \begin{array}{ccc} 1 & 0 &...
Iosif Pinelis's user avatar
6 votes
Accepted

Image of the trace map of ring of integers

Question 1. Yes, and this follows from the results of Chapter VIII in Weil: Basic Number Theory. See especially Corollary 2 of Proposition 4 in that chapter. Question 2. In general, the exponent of $...
GH from MO's user avatar
  • 101k
6 votes
Accepted

Which operators on the trace-class operators extend to operators on Hilbert-Schmidt operators?

No, this space is not closed under involution. Choose $A \in TC$ and $B \in HS\setminus TC$ and consider the rank one operator $T \mapsto \langle T, B\rangle A$ on $HS$. This restricts to a bounded ...
Nik Weaver's user avatar
  • 42.4k
5 votes
Accepted

Trace-class operator satisfies $\sum |\lambda_n|<\infty$?

As I mentioned in a comment, exercise 3 has a positive answer: Nuclear operators are absolutely $2$-summing, and $2$-summing operators have $2$-summable eigenvalues (see, e.g., Tomczak's book). ...
Bill Johnson's user avatar
5 votes

Reconstruct a matrix from its traces

I do not think you can reconstruct $A$ just from this information. Take $\Gamma=I_n$, let $M$ be any orthogonal $n \times n$ matrix and set $B={}^tMA M$. Then $A$ and $B$ are two similar (symmetric)...
Francesco Polizzi's user avatar
5 votes
Accepted

Algorithm to minimize $\operatorname{tr}(PAP^TB)$?

This problem is NP-hard. Let $A$ be the adjacency matrix of an $n$-cycle plus the all-ones matrix and $B$ the adjacency matrix of a graph $G$ plus the all-ones matrix. Then, if $G$ has a Hamiltonian ...
LeechLattice's user avatar
  • 9,431
5 votes
Accepted

Hattori-Stallings trace

The answer to (1) is yes. In fact this is true more generally: for any $f: M\to N, g:N\to M$, you have $tr(fg) = tr(gf)$. The point is that $\hom_R(M,N)\otimes \hom_R(N,M)\cong \hom_R(M,R)\otimes_R N \...
Maxime Ramzi's user avatar
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4 votes
Accepted

Retractions for completely positive unital maps, with particularly nice norms

The answer is no again. When $h=k$, non-singular is the same as bijective so any $\Psi$ has only one map $\Phi$ such that $\Phi\circ\Psi = I_{M_k}$, namely the inverse. In the example below we find a ...
Chris Ramsey's user avatar
  • 3,954
4 votes

Submodularity property of trace of inverse matrix

This submodularity property is known to not hold, as the OP has found out already. However, I'd like to mention the following observation: Let $f$ be defined on $(0,\infty)$ such that $-f'$ is ...
Suvrit's user avatar
  • 28.4k
4 votes

Norm and trace inequalities

As Christian said, no. Your condition ``$A$ and $B$ are two positive definite matrices such that $\|A\| \leq \|B\|$ for every unitarily invariant norm $\| \cdot \|$" is equivalent to "$\max\limits_{VU ...
M. Lin's user avatar
  • 1,738
4 votes

Trace of six gamma matrices

Unfortunately, I can not comment on other answers, hence the reply. I do not agree with @Zurab Silagadze. The identity $$\gamma_\mu \gamma_\nu \gamma_\rho = g_{\mu \nu} \gamma_\rho - g_{\mu \rho} \...
Marvin's user avatar
  • 41
4 votes

Geometric interpretation of trace

The trace of a linear map $A : V \to V$ is always the same as the trace of its restriction to the unique largest vector subspace $W$ of $V$ such that $A\big\vert_W : W \to W$ is an isomorphism. By ...

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