Bounded operators and adjoints
Control linear amplification and define the adjoint without relying on a matrix transpose.
On this page
The question for today
What you’ll understand
- Relate boundedness to continuity for an everywhere-defined linear operator.
- Compute an adjoint using the specified inner product.
- Explain how an operator norm controls errors and compositions.
A linear rule can amplify small changes by different amounts in different directions. Its operator norm captures the worst amplification over all unit inputs. In finite dimensions this is always finite, but an infinite list of finite-looking coordinate rules can have no uniform bound. Adjoint operators transfer a linear action from one side of an inner product to the other. This makes them central to normal equations, observables, sensitivity calculations, and the geometry of the range.
A bound that works in every direction
For a linear map T defined on all of a normed space H and taking values in another normed space K, bounded means there is one finite constant C with norm(Tx) at most C norm(x) for every x. The smallest such constant is the operator norm. Linearity turns continuity at zero into this uniform inequality, and the inequality implies continuity everywhere. Merely showing that each individual output has finite norm does not establish one constant that works for all inputs.
On square-summable sequences, the rule that multiplies coordinate n by 1/n is bounded with norm one. The rule multiplying coordinate n by n is not an everywhere-defined map from that space into itself: some square-summable inputs acquire nonsummable outputs. It can instead define an unbounded operator on a proper domain. Keeping that distinction prevents an invalid appeal to the closed graph theorem, which requires an everywhere-defined map between Banach spaces with closed graph.
The adjoint follows the inner product
For a bounded map T:H→K between Hilbert spaces, fix y in K. The expression inner-product(y,Tx) is a bounded linear functional of x under our convention. Riesz representation gives a unique vector T* y in H representing that functional. This defines an everywhere-defined bounded map T*:K→H. The defining identity is intrinsic, so the adjoint exists even without a coordinate basis or a finite matrix representation.
In orthonormal coordinates the adjoint matrix is the conjugate transpose. If coordinates carry a weighted inner product, blindly taking a transpose is wrong: the weight matrices enter the formula. Over real spaces no conjugation is needed, but the weights still matter. A bounded operator equals its adjoint when it is self-adjoint. A bounded operator is normal when it commutes with its adjoint. Neither property means that its matrix is diagonal in the coordinates currently displayed.
Ranges, kernels, and least squares
A vector is perpendicular to the range of T exactly when it belongs to the kernel of T*. Consequently the closure of the range of T is the orthogonal complement of that kernel. The closure cannot always be dropped: the diagonal operator with entries 1/n has dense but nonclosed range on square-summable sequences. Inverse problems can therefore have outputs arbitrarily close to attainable data even when the requested exact solution does not lie in the allowed input space.
For finite least squares, differentiating the squared residual gives the normal equation T*T x=T*b. In infinite dimensions a minimizer may fail to exist if the range is not closed. Even in finite dimensions, forming T*T squares a full-column-rank matrix's Euclidean condition number, so an identity useful for reasoning may be a poor numerical algorithm. Error amplification, solvability, and a stable computational method are related questions that still require separate checks.
Work it through
Check a shear and its adjoint
With the usual real inner product, take , x=(1,−1), and y=(2,3). Verify the adjoint identity.
Compute Tx=(−1,−1), so y·Tx=−2−3=−5.
The adjoint is the transpose, giving T*y=(2,7). Therefore (T*y)·x=2−7=−5, matching the first calculation.
For the test input e_2=(0,1), the output is (2,1), with norm √5. Thus the operator norm is at least √5; this single test is a lower bound, not proof that the supremum equals √5.
The pairing agrees exactly at −5. A coordinate example verifies the definition while exposing the distinction between one-direction amplification and the full operator norm.
The living notebook
Explore the model
Interactive lab · vectors
A norm is independent of direction
Hold magnitude at 1 and sweep the angle. The coordinate components change while the input norm stays fixed. Use this to understand the unit-input set over which an operator norm takes its supremum. This lab does not apply T, compute its singular values, or display its adjoint.
- x component
- 1.393
- y component
- 0.975
- Magnitude
- 1.7
Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA worst-case bound must cover every unit direction; one favorable input is insufficient.
Pause & reason
Check your understanding
Your turn
Try it yourself
For the real diagonal matrix D=diag(2,−3), compute its adjoint and Euclidean operator norm. Give an input achieving that norm.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Compare the squared output norm with 9 times the squared input norm.
Compare with the solution
D*=D. For x=(a,b), the squared output norm is 4a²+9b²≤9(a²+b²), so the operator norm is at most 3. The unit input (0,1) has output norm 3, achieving the bound. The negative eigenvalue changes direction but its absolute value determines amplification.
Keep the idea
Take it with you
- Boundedness is a uniform linear amplification bound, equivalent to continuity.
- The adjoint depends on the inner products and reverses composition.
- Closedness of the range matters for solvability and least-squares approximation.
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