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 ...

Community wiki

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 ...

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}^...

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 ...

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^...

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 ...

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 ...

Community wiki

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 ...

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 ...

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 $M_n$ ...

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}|...

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}...

Community wiki

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-...

Community wiki

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) =...

8
votes

Accepted

### Is there a converse to the Brauerâ€“Nesbitt theorem?

Not always â€” 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 (...

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 ...

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 ...

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 $\{\...

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 &...

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 $...

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 ...

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).
...

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)...

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 ...

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 \...

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 ...

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 ...

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 ...

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} \...

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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