Linear Algebra - Lecture 18 - Fall, 2026
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Overview
The lecture proves two vector-space results, then develops real inner product spaces through dot products, norms, orthogonality, and orthonormal bases. It derives Bessel’s inequality and Cauchy–Schwarz, uses Cauchy–Schwarz to prove the triangle inequality, and closes with the formula and uniqueness of the projection of a vector onto a given direction.
Key takeaways
- For a linear operator T, the condition T² = 0 is equivalent to im(T) ⊆ ker(T), because every output T(v) must be sent to zero by a second application of T.
- The fixed-point set {v : T(v) = v} is a subspace: linearity preserves the fixed-point equation under vector addition and scalar multiplication.
- An orthonormal set lets each coefficient be computed directly as cᵢ = ⟨v,eᵢ⟩, and Bessel’s inequality guarantees Σcᵢ² ≤ ‖v‖² even if the set is not a full basis.
- Cauchy–Schwarz follows by requiring ‖tu+v‖² to be nonnegative for every real t; its quadratic discriminant must therefore be nonpositive.
- The triangle inequality follows by expanding ‖u+v‖² and bounding ⟨u,v⟩ with Cauchy–Schwarz; when u and v are orthogonal, the same expansion gives the Pythagorean identity.
- For nonzero w, v decomposes uniquely into (⟨v,w⟩/‖w‖²)w plus a vector perpendicular to w, providing the geometric foundation for Gram–Schmidt.
Chapters
- For a linear operator T, T²(v) means T(T(v)) for every vector v.
- T² = 0 exactly when every T(v) lies in ker(T), which is equivalent to im(T) being contained in ker(T).
- A fixed point satisfies T(v) = v; the set W of all fixed points contains 0 because T(0) = 0.
- Linearity shows W is closed under addition and scalar multiplication: if v₁, v₂ ∈ W, then T(v₁ + v₂) = v₁ + v₂, and T(cv₁) = cv₁.
- In Rⁿ, the standard inner product is ⟨x,y⟩ = Σxᵢyᵢ, the sum of products of corresponding coordinates.
- In R³, (2, 7, −4) and (6, 0, 3) have inner product 12 − 12 = 0, so they are perpendicular.
- In R², (4, 5) and (−10, 8) have inner product −40 + 40 = 0, another example of perpendicular vectors.
- An inner product on a real vector space is bilinear, meaning linear separately in each argument.
- Symmetry requires ⟨v,w⟩ = ⟨w,v⟩, while positive definiteness requires ⟨v,v⟩ > 0 for every nonzero v.
- The norm is defined by ‖v‖ = √⟨v,v⟩; for v = (2, 7, −4) in R³, ‖v‖ = √69.
- The real polarization identity recovers the inner product from norms: ⟨v,w⟩ = ¼(‖v+w‖² − ‖v−w‖²).
- An orthogonal set contains nonzero vectors with pairwise inner products 0; a unit vector has norm 1.
- Every nonzero vector v can be normalized as v/‖v‖, which has norm 1.
- Vectors are parallel when one is a nonzero scalar multiple of the other; a negative scalar reverses direction.
- An orthonormal basis consists of unit vectors that are pairwise orthogonal, so coordinates xᵢ and yᵢ give ⟨v,w⟩ = Σxᵢyᵢ.
- The Gram–Schmidt orthonormalization algorithm converts a basis into an orthonormal basis; its procedure is reserved for the next lecture.
- For an orthonormal set e₁,…,eₙ, the coefficient of v along eᵢ is cᵢ = ⟨v,eᵢ⟩.
- These coefficients are also called Fourier coefficients, connecting finite-dimensional inner product calculations to Fourier series.
- The lecture introduces Bessel’s inequality for a finite orthonormal set, even when that set is not a basis.
- Set v′ = v − Σcᵢeᵢ, where cᵢ = ⟨v,eᵢ⟩ for the orthonormal set.
- Positive definiteness gives ⟨v′,v′⟩ ≥ 0; expanding and using ⟨eᵢ,eⱼ⟩ = 0 for i ≠ j yields ‖v‖² − Σcᵢ² ≥ 0.
- Therefore, the sum of squared Fourier coefficients is at most ‖v‖², with equality when the residual is zero.
- For a real quadratic at² + bt + c, the discriminant b² − 4ac determines whether it has zero, one, or two real roots.
- The Cauchy–Schwarz inequality for an inner product space is |⟨u,v⟩| ≤ ‖u‖‖v‖.
- The proof uses the fact that ‖tu + v‖² is nonnegative for every real t, so the resulting quadratic cannot have two distinct real roots.
- Expanding gives ‖tu + v‖² = t²‖u‖² + 2t⟨u,v⟩ + ‖v‖², using symmetry of the real inner product.
- Because this quadratic is nonnegative for every real t, its discriminant is at most zero.
- Rearranging gives ⟨u,v⟩² ≤ ‖u‖²‖v‖²; taking square roots yields |⟨u,v⟩| ≤ ‖u‖‖v‖.
- Apply the inner product result to a = (a₁,…,aₙ) and b = (b₁,…,bₙ) in standard Rⁿ to obtain |Σaᵢbᵢ| ≤ √(Σaᵢ²)√(Σbᵢ²).
- For a = (1, 2, 3) and b = (4, 5, 6), the dot product is 32, while the norm product is √14√77 ≈ 32.81.
- The numerical example confirms that the absolute dot product does not exceed the product of the vector lengths.
- If ⟨u,v⟩ = 0, expansion gives ‖u+v‖² = ‖u‖² + ‖v‖², the Pythagorean theorem for orthogonal vectors.
- For arbitrary u and v, Cauchy–Schwarz bounds the cross term in ‖u+v‖², giving ‖u+v‖² ≤ (‖u‖ + ‖v‖)².
- Taking square roots yields the triangle inequality: ‖u+v‖ ≤ ‖u‖ + ‖v‖.
- For nonzero w, the component of v parallel to w is proj_w(v) = (⟨v,w⟩/‖w‖²)w; the remainder v − proj_w(v) is perpendicular to w.
- If v has two decompositions into a component parallel to w and a component perpendicular to w, subtracting them makes a vector both parallel and perpendicular to w.
- Since ‖w‖² > 0, that parallel difference must be zero; both components are therefore unique, a fact used in Gram–Schmidt.
Transcript excerpt
0:00 Okay, lecture 18 in linear algebra and I was asked to do two problems from the chapter on vector spaces. Um, so problem number 14 says the following. We have t from a vector space to itself which is a linear operator. linear operator is simply a linear transformation but from one vector space to itself. And we want to prove that t ^2 is zero if and only if the image of t is contained in the kernel of t. So proof.
0:50 So recall t ^2 of a vector v means t of t of v, right? And t ^2 of v which is t of t of v equals zero means what? If and only if t of v. So t sends t of v to zero. So t of v is in the kernel of t. So let's say t ^2 is zero if and only if t ^2 of v is zero for all vectors v in the vector space. So if and only if t of
1:38 v is in the kernel of t for all vectors v in the vector space. But the set of all T of V is the image of T is contained in the kernel of T. Right? So if the image of T is contained in the kernel of T, then for every vector V, T of V is in the kernel. So T ^2 of V is zero. And that's the entire proof. If t^ square of v is zero then the image is contained in the kernel and if the image is in the kernel then t^2 is zero and any questions about that
2:29 in problem 15 again we have a linear operator t a fixed point of t so a vector V is a fixed point of T means T of V is
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