Matrix Analysis

Norm

Norm for vectors

\[\begin{aligned} &\|x\| \geq 0 \\ &\|x\|=0 \text { if and only if } x=0 \\ &\|c x\|=|c|\|x\| \\ &\|x+y\| \leq\|x\|+\|y\| \end{aligned} \]

sum norm(\(l_1\)-norm)

\[\|x\|_{1}=\left|x_{1}\right|+\cdots+\left|x_{n}\right| \]

Euclidean norm(\(l_2\)-norm)

\[\|x\|_{2}=\left(\left|x_{1}\right|^{2}+\cdots+\left|x_{n}\right|^{2}\right)^{1 / 2} \]

max norm(\(l_\infty\)-norm)

\[\|x\|_{\infty}=\max \left\{\left|x_{1}\right|, \ldots,\left|x_{n}\right|\right\} \]

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Norm for matrix

\(l_1\)-norm

\[\|A\|_{1}=\sum_{i, j=1}^{n}\left|a_{i j}\right| \]

\(l_2\)-norm,Frobenius norm,Schur norm, Hilbert-Schmidt norm

\[\|A\|_{2}=\left|\operatorname{tr} A A^{*}\right|^{1 / 2}=\left(\sum_{i, j=1}^{n}\left|a_{i j}\right|^{2}\right)^{1 / 2} \]

\(l_\infty\)-norm

\[\|A\|_{\infty}=\max _{1 \leq i, j \leq n}\left|a_{i j}\right| \]

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More about norms

Operator norm^[1]: \(||T||\equiv sup_{||v||=1}||T v||\). If the \(||\cdot||\) stands for \(l_2\)-norm, then utilizing \(l_2\) norm can be written as \(||a||=\sqrt{a^\dagger a}\), we can write the operator norm as \(\sqrt{v^\dagger T^\dagger Tv}\), the max value of the term under the square root is the maximal eigenvalue of \(T^\dagger T\) using the theorem of maximal expectation value of hermitian matrix. So the operator norm is the maximal of the singular value of the operator \(T\).

Trotter formulas or Trotter–Suzuki decompositions

If \({\displaystyle H=A+B+C}{\displaystyle H=A+B+C}\), then \({\displaystyle U=e^{-i(A+B+C)t}=(e^{-iAt/r}e^{-iBt/r}e^{-iCt/r})^{r}}{\displaystyle U=e^{-i(A+B+C)t}=(e^{-iAt/r}e^{-iBt/r}e^{-iCt/r})^{r}}\) for a large \({\displaystyle r}\).
Taylor expansion \(e^{-iAt/r}\) term, find that \(e^{-iAt/r}e^{-iBt/r}e^{-iCt/r}\approx e^{-i(A+B+C)t/r}\). \(\blacksquare\)

Visualization of Matrix inequality

A matrix can be visualized as an ellipsoid. Then inequality \(A\ge B\) is equal to ellipsoid of \(A\) contain the ellipsoid of \(B\).

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References

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1. https://mathworld.wolfram.com/OperatorNorm.html ↩︎