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 \(T\).

References

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