Operator norm

The operator norm is $\Vert T\Vert ={\sup }_{\Vert x\Vert \le 1}\Vert Tx\Vert$.

For bounded operators between Hilbert spaces, write the norm as the supremum of $|⟨Ax,y⟩|$ over unit vectors $x,y$. Moving the operator to the other slot proves equality of its norm with that of its adjoint operator. Applied to the exponential matrix, this makes the primal and dual analytic large sieve inequalities equivalent.

The Euclidean logarithmic norm is the largest eigenvalue of a matrix's Hermitian part. Differentiating $\Vert y{\Vert }_2^2$ along $y^{\prime }=Ay$ gives $\Vert e^{tA}{\Vert }_2\le e^{t{\mu }_2(A)}$. It is the least exponent for such a bound with prefactor one, as a first-order expansion at zero shows. It equals the spectral abscissa for a normal matrix, but need not do so otherwise. A scalar rational stability function does not generally inherit a bound by its value at this real number.

For composable bounded linear operators, the operator norm satisfies $\Vert AB\Vert \le \Vert A\Vert \Vert B\Vert$. Indeed, $\Vert ABx\Vert \le \Vert A\Vert \Vert Bx\Vert \le \Vert A\Vert \Vert B\Vert \Vert x\Vert$, and taking the supremum over unit vectors proves the bound. It implies $\Vert [A,B]\Vert \le 2\Vert A\Vert \Vert B\Vert$ by the triangle inequality.

If $A_j$ and $B_j$ have operator norm at most one, then inserting and subtracting one factor at a time gives

Insert one factor difference at a time: the $j$th term is $U_m\cdots U_{j+1}(U_j-V_j)V_{j-1}\cdots V_1$. Multiplication by the surrounding unitary operators preserves the spectral norm, so the triangle inequality gives the displayed dimension-independent bound. No commutation assumption is required.

The matrix 2-norm is the operator norm induced by the Euclidean norm:It equals the largest singular value of $A$ and is invariant under multiplication by orthogonal or unitary matrices.