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Results tagged with operator-norms
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user 8430
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, $$|\text{trace}(AB)| \le \sum_{i=1} …
- 28.6k
10
votes
Accepted
Norm estimation of identity plus two non-commuting self-adjoint operators
The claim is false. Consider the following matrix argument.
\begin{eqnarray*}
\|(I+A+B)^{-1}A\| \le 1\quad\Leftrightarrow\quad \begin{bmatrix}I & (I+A+B)^{-1}A \\ A(I+A+B)^{-1} & I \end{bmatrix} \ge …
- 28.6k
10
votes
Accepted
Bounding the matrix norm of a commutator $[A,B]$ in terms of the norms of $A$ and $B$
A somewhat more general setting, namely, finding the best constant $C_{p,q,r}$ in
\begin{equation*}
\|AB-BA\|_p \le C_{p,q,r}\|A\|_q\|B\|_r,
\end{equation*}
for Schatten $p$,$q$,$r$-norms, is studied …
- 28.6k
9
votes
Bounding the trace of a matrix product by the operator norms; generalized Hölder inequality?
I include some information about Hölder's inequality just for completion of details for Mikael's nice answer.
The Schatten-$p$ norm of a matrix $X$ is defined as
$$\lVert X\rVert_p := \Bigl(\sum\nolim …
- 28.6k
4
votes
$\|T\|_2 \le \sqrt{\|T\|_1\|T\|_\infty}$
Sorry, my answer below is only partial, but I thought that it may still be somewhat interesting.
As far as I know, this inequality does not have a distinguished name. It is ultimately a consequence …
- 28.6k