27
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 ...
- 6,406
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, $$|\text{trace}(AB)| \le \sum_{i=1}^...
- 28.6k
15
votes
Accepted
Trace of non-commutable matrices
Your conjecture is a special case of the following result which essentially follows from the Lieb-Thirring inequality.
Let $A$ and $B$ be Hermitian matrices. Then, for every positive integer $p$ we ...
- 28.6k
12
votes
Accepted
Reconstruct a matrix from its traces
Unfortunately even the generic situation is bad. Since we know the eigenvalues, we should search for the orthonormal system of eigenvectors $v_i$ of $A$. We have ($e_i$ is the standard basis)
$$
Tr(A^...
- 61.9k
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
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 ...
- 8,529
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 ...
- 188k
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 ...
- 96.4k
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 $M_n$ ...
- 42.8k
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
$$
tr(AB) \leq|tr(AB)|=\left|\sum a_{ii}b_{ii}\right| \leq \sum a_{ii}|...
- 2,361
8
votes
Geometric interpretation of trace
Was surprised not to see this here yet. Let $V$ be an $n$-dimensional real vector space with inner product.
Any linear transformation $f:V \to V$ can be decomposed into
$$ f = \left(\tfrac{\textrm{tr}...
8
votes
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
$$\Phi\left(\left[\begin{array}{cc} A & B\\ C& D\end{array}\right]\right) =...
- 3,984
8
votes
Accepted
Is there a converse to the Brauer–Nesbitt theorem?
Not always — e.g. $g(1)$ should be an integer. The desired description is given in
Helling, H., Eine Kennzeichnung von Charakteren auf Gruppen und assoziativen Algebren, Commun. Algebra 1, 491-501 (...
- 31.5k
8
votes
If I multiply the coefficients of a trace-class operator with bounded complex numbers is it still trace class?
Nik Weaver's answer gives a nice counter-example. Let me just say a few words of context. Kernels $K$ for which $KT$ is trace-class for all trace-class $T$ are called Schur multipliers. (Not to be ...
- 18.7k
7
votes
Accepted
Bound on sum of $n$th super-diagonal entries in a $2n$ by $2n$ PSD matrix
If $\left(\begin{array}{} A & B \\ B^T & C \end{array}\right)\succeq 0$ then there exists a contraction $K$ (i.e. $\lambda_n(K)=\|K\| \leq 1$) such that $B = A^{1/2} K C^{1/2}$ (e.g. see ...
- 1,991
7
votes
Inequality for trace of a symmetric product?
If the $v_i$ are orthonormal, then yes. If the $v_i$ aren't assumed to be orthogonal, they can cluster around the dominant eigenvector and cause a counterexample.
Let the eigenvalues of $A$ be $\{\...
- 1,593
7
votes
Trace inequality under consideration of definiteness
Write
$$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 &...
- 128k
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 $...
- 105k
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 ...
- 42.8k
6
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 \...
- 15.7k
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).
...
- 31.5k
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)...
- 66.3k
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 ...
- 9,501
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 ...
- 28.6k
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 ...
- 1,748
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 ...
- 3,984
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} \...
- 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 ...
4
votes
Geometric interpretation of trace
For me trace of a matrix always was analogous to the real part of a complex number. As such, I consider trace divided by the matrix' rank as the "scalar part" of the matrix.
There are other ...
Only top scored, non community-wiki answers of a minimum length are eligible