4.5.33 · D5Linear Algebra (Full)
Question bank — Inner product spaces — dot product generalization
The three axioms we keep returning to: symmetry, linearity in the first slot, and positive-definiteness ( with equality only for ). Almost every trap here is really "which axiom did you forget?"
True or false — justify
True or false: any symmetric bilinear form on a real space is an inner product.
False — it must also be positive-definite. is symmetric and bilinear but , so lengths would be imaginary.
True or false: if then need not be the zero vector.
False for a genuine inner product — positive-definiteness forces . If some nonzero gives , the form is only "positive-semidefinite", not an inner product.
True or false: two vectors can be orthogonal under one inner product but not another.
True — orthogonality means , and that number depends on the chosen inner product. E.g. and are orthogonal under but not under .
True or false: the zero vector is orthogonal to every vector.
True — by linearity for all . The zero vector is the one vector orthogonal to everything, including itself.
True or false: can only happen when .
False — equality in the triangle inequality holds when point the same way ( with ), not necessarily equal. The cross term must hit its ceiling .
True or false: negating a weight in still gives an inner product.
False — the negative weight kills positive-definiteness: . Every weight in a weighted inner product must be strictly positive.
True or false: Cauchy–Schwarz is a theorem you must prove, not an axiom you assume.
True — it is derived from positive-definiteness via . It is the price the axioms pay so that lands in . See Cauchy-Schwarz Inequality.
True or false: if for all , then .
True — in particular take , giving , so positive-definiteness forces . Only the zero vector is orthogonal to the whole space.
Spot the error
Spot the error: "I computed , so the angle is just a bit unusual."
A cosine can never exceed ; Cauchy–Schwarz guarantees . A value means an arithmetic slip or a form that isn't a real inner product.
Spot the error: "Symmetry lets me write , so I can pull from either slot freely."
The conclusion is true (for real inner products) but the reason is linearity plus symmetry, not symmetry alone. Symmetry only swaps slots; linearity is what scales.
Spot the error: " on degree- polynomials is a valid inner product."
It fails positive-definiteness: any constant nonzero has , so while . A whole line of nonzero vectors has "zero length".
Spot the error: "Since , these functions are literally at on the graph."
They are orthogonal in the integral inner product, which is the abstract meaning of "" here — it is not a visible right angle between curves. Orthogonality lives in the inner product, not on the page.
Spot the error: "The distance because subtraction measures how far apart the lengths are."
Distance is , the norm of the difference vector, not the difference of norms. Two vectors of equal length are far apart if they point differently. See Norms and Distance.
Spot the error: "In Gram–Schmidt I can normalize a vector even if ."
You cannot — dividing by is undefined. That situation never arises for a nonzero under a real inner product, precisely because of positive-definiteness. See Orthogonality and Gram-Schmidt.
Why questions
Why must the discriminant in the Cauchy–Schwarz proof be , not ?
The quadratic is (never strictly below zero), so it may touch the axis — that gives and the equality case where are parallel.
Why does positive-definiteness, not symmetry, guarantee real lengths?
Length is ; only positivity () keeps the number under the root non-negative so the square root is real.
Why is the integral a natural inner product for functions?
It mimics the Dot Product by summing over a continuum of "coordinates" , and it satisfies all three axioms, giving functions genuine length, angle, and orthogonality — the foundation of Fourier Series.
Why does the triangle inequality need Cauchy–Schwarz?
Expanding produces a cross term ; only Cauchy–Schwarz bounds it above by , which is exactly what closes the square .
Why do we insist linearity holds in the first slot specifically?
For a real inner product, symmetry then hands us linearity in the second slot for free, so one slot suffices as an axiom. Stating both would be redundant.
Why is orthogonal projection only well-defined once we have an inner product?
Projection needs a notion of "the component of along ", which is — both dots require an inner product to even be numbers. See Orthogonal Projections.
Edge cases
Edge case: is the only way two nonzero vectors give equality in Cauchy–Schwarz?
No — equality happens when they are parallel (linearly dependent), the opposite extreme from orthogonal where the value is .
Edge case: what is the angle between a vector and itself?
, so — a nonzero vector is perfectly aligned with itself, exactly as intuition demands.
Edge case: can equal exactly ?
Yes, when with (anti-parallel). Cauchy–Schwarz is tight at both ends , corresponding to same-direction and opposite-direction vectors.
Edge case: on the zero vector space , is there an inner product?
Trivially yes — the only pair is , which satisfies all axioms vacuously, including positive-definiteness since is the zero vector.
Edge case: what happens to if one vector is ?
It is undefined — the formula divides by . The zero vector has no direction, so "angle to it" is meaningless, even though makes it orthogonal to all.
Edge case: does a weighted inner product change which vectors are orthogonal?
Yes — under , the pair gives (orthogonal), while the standard dot product gives . Weights reshape the geometry.
Recall One-line summary of every trap
Nearly every "concept trap" here is a disguised failure of positive-definiteness, a misuse of Cauchy–Schwarz (angles must stay in ), or forgetting that orthogonality is relative to the chosen inner product. When stuck, check those three first.