Operator norm

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

For a linear operator bounded from $L^{p_i}$ to $L^{q_i}$ with norms $M_i$, $i=0,1$, interpolate reciprocals by $1/p=(1-\theta )/p_0+\theta /p_1$ and $1/q=(1-\theta )/q_0+\theta /q_1$. The intermediate operator norm is at most $M_0^{1-\theta }{M}_1^{\theta }$. The complex interpolation proof first treats simple functions; completeness and continuity on the endpoint sum spaces identify the extension with the original operator. On finite counting spaces it also gives norm bounds between sequence spaces.

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.

The spectral radius of the matrix exponential gives the lower bound through the spectral abscissa $s(A)$. Differentiating the squared Euclidean norm along $q^{\prime }=Aq$ and bounding its Hermitian quadratic form gives the upper bound through the Euclidean logarithmic norm ${\mu }_2(A)$. A non-normal matrix can have ${\mu }_2(A)>0$ despite $s(A)<0$, allowing transient growth before its ultimate decay.

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.

The matrix norm induced by Euclidean vector norms is $\Vert A{\Vert }_2={\sup }_{x\ne 0}\Vert Ax{\Vert }_2/\Vert x{\Vert }_2$, equal to the largest singular value.