A linear operator T: ℝ² → ℝ² is represented by a matrix whose largest singular value is 5. With respect to the standard Euclidean norm, what is ‖T‖ and what does it mean geometrically?
A25 — the operator norm squares the largest singular value to account for area distortion
B5 — the operator norm equals the largest singular value, measuring the maximum stretching of any unit vector
C√5 — singular values appear under a square root in the norm computation
DThe operator norm cannot be determined from singular values without knowing T's nullspace
For a matrix with standard Euclidean norms, ‖T‖ equals its largest singular value. Geometrically, T maps the unit sphere to an ellipsoid; the largest singular value is the length of the longest axis of that ellipsoid — the worst-case stretching factor. The operator norm answers 'by at most how much can T amplify a vector?' The answer is σ₁ = 5 directly, with no squaring or square-rooting. This worst-case amplification is exactly what the definition sup{‖T(x)‖ : ‖x‖ ≤ 1} computes.
Question 2 Multiple Choice
Suppose ‖S‖ = 3 and ‖T‖ = 4. A student calculates ‖ST‖ = 12, claiming the composition stretches exactly as much as the two operators combined. What is the correct statement?
AThe student is correct: ‖ST‖ = ‖S‖‖T‖ = 12 always holds for composed operators
BSubmultiplicativity gives only an upper bound: ‖ST‖ ≤ 12, and the actual norm could be strictly less
CComposition reverses the order, so the student should compute ‖TS‖ = 12 instead
DSubmultiplicativity only applies to self-adjoint operators, not arbitrary S and T
Submultiplicativity states ‖ST‖ ≤ ‖S‖‖T‖ — an inequality, not an equality. The composition ST may stretch less than 12 because S might contract in the direction that T stretched: for example, if T expands along the x-axis and S contracts along the x-axis, their composition could have a much smaller norm. Equality holds in special cases (e.g., both diagonal with aligned extremal directions) but not in general. The inequality is all that is guaranteed — and it is exactly what is needed for ℒ(X,X) to be a Banach algebra.
Question 3 True / False
‖T‖ = 0 if and only if T is the zero operator.
TTrue
FFalse
Answer: True
This is the definiteness axiom of a norm applied to ℒ(X,Y). If ‖T‖ = 0, then sup{‖T(x)‖_Y : ‖x‖_X ≤ 1} = 0, so ‖T(x)‖_Y = 0 for every x in the unit ball, hence for every x by linearity (scaling x to lie in the unit ball). Therefore T(x) = 0 for all x, meaning T is the zero operator. The converse is immediate. This verification — along with the triangle inequality and homogeneity — is what makes ‖·‖ a genuine norm on ℒ(X,Y).
Question 4 True / False
If Y is a Banach space, then ℒ(X,Y) is also a Banach space for any normed space X, because completeness of the codomain is sufficient for completeness of the operator space.
TTrue
FFalse
Answer: True
This is the theorem: ℒ(X,Y) is Banach whenever Y is Banach, regardless of whether X is complete. The key is that Cauchy sequences of operators {Tₙ} in ℒ(X,Y) produce, for each fixed x, Cauchy sequences {Tₙ(x)} in Y. Since Y is complete, these converge to some T(x) ∈ Y. One then verifies that T is linear and bounded, and that Tₙ → T in operator norm. The completeness of X plays no role in this argument — what matters is that the values Tₙ(x) live in a complete space.
Question 5 Short Answer
Why is the operator norm defined as a supremum over the unit ball, and what does this reveal about the relationship between an operator being bounded and having a finite operator norm?
Think about your answer, then reveal below.
Model answer: The supremum over the unit ball efficiently captures worst-case stretching because linear operators scale predictably: ‖T(αx)‖ = |α|‖T(x)‖, so the 'worst direction' is the same at every scale. By normalizing to unit vectors, we extract a scale-free amplification factor. An operator is bounded precisely when this supremum is finite — i.e., when there exists a uniform constant C with ‖T(x)‖ ≤ C‖x‖ for all x. The operator norm is the tightest such C: the infimum of all valid constants, realized as the actual supremum. Boundedness and finite operator norm are equivalent characterizations of the same property.
The equivalence ‖T‖ = sup{‖T(x)‖/‖x‖ : x ≠ 0} makes the 'worst-case ratio' interpretation explicit. The operator norm is not just any bound — it is the sharp bound, the actual maximum amplification achieved (or approached) by T.