# Tag Info

### 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 ...

### 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 ...
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### 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 ...
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### 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 ...

### 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 ...
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### 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 ...
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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 $M_n$ ...
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### 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-...
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### 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 ...
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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 ...