Practice: Abstract vector spaces
Self-check
Section titled “Self-check”Six short questions. Answer each one in your head (or on paper) before opening the collapsible. Trying to retrieve the answer is where the learning sticks; rereading feels productive but does much less.
1. What is the actual definition of a vector (the math view from the first lesson)?
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Anything you can add and scale coherently: you can add two of them and get another of the same kind, and scale one by a number and get another of the same kind. Arrows and lists are two examples; they were the teaching model, not the definition.
2. Are functions vectors? Why?
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Yes. You can add two functions pointwise ((f + g)(x) = f(x) + g(x)) and scale one pointwise ((c · f)(x) = c · f(x)), and both stay inside the set of functions. Addition and scaling behave coherently, so by the math view’s test, functions form a vector space.
3. What is a basis for polynomials of degree 3 or less, and what is the dimension?
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The basis is {1, x, x^2, x^3}: every such polynomial is a unique linear combination of these four pieces. There are four of them, so the space has dimension 4.
4. How does a polynomial get coordinates?
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By fixing a basis. In {1, x, x^2, x^3}, the polynomial 2x^2 + 5x + 7 is 7·1 + 5·x + 2·x^2 + 0·x^3, so its coordinate vector is [7, 5, 2, 0]. It then behaves like any vector: adding polynomials is adding their coordinate vectors.
5. How is the derivative a matrix?
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Differentiation is linear (the derivative of a sum is the sum of derivatives, and of a scaled function is the scaled derivative), so it is captured by where it sends the basis. Recording d/dx of each basis polynomial as a column gives the derivative matrix, and differentiating becomes matrix-vector multiplication.
6. Does abstraction introduce new rules?
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No. Functions and polynomials use the same addition, scaling, matrices, and eigenvectors as arrows. The objects changed; the algebra did not. Any set whose addition and scaling obey the standard axioms is a vector space, and every tool from the whole track applies to it.
Try it yourself, part 1: coordinates, addition, and differentiation
Section titled “Try it yourself, part 1: coordinates, addition, and differentiation”Work in the basis {1, x, x^2, x^3} with the derivative matrix D = [[0,1,0,0], [0,0,2,0], [0,0,0,3], [0,0,0,0]]. About 8 minutes, pen and paper.
Step 1. Write 3x^3 + x^2 + 4 as a coordinate vector.
Step 2. Add (x^3 + 2x) and (x^2 + 5) by adding their coordinate vectors, then read the polynomial back.
Step 3. Differentiate 3x^3 + x^2 + 4 by computing D times its coordinate vector, read the result back as a polynomial, and check it against ordinary calculus.
Check your work
Step 1. Constant 4, no x, one x^2, three x^3: [4, 0, 1, 3].
Step 2. x^3 + 2x is [0, 2, 0, 1] and x^2 + 5 is [5, 0, 1, 0]. Sum: [5, 2, 1, 1], which reads back as x^3 + x^2 + 2x + 5.
Step 3. D · [4, 0, 1, 3] = 4·[0,0,0,0] + 0·[1,0,0,0] + 1·[0,2,0,0] + 3·[0,0,3,0] = [0, 2, 9, 0], which reads back as 9x^2 + 2x. Ordinary calculus: d/dx(3x^3 + x^2 + 4) = 9x^2 + 2x. They match: calculus by matrix multiplication.
Try it yourself, part 2: eigenvectors of the derivative
Section titled “Try it yourself, part 2: eigenvectors of the derivative”On the space of all functions, the derivative d/dx is a linear transformation, so it has eigenvectors. About 6 minutes.
Step 1. Confirm that e^(3x) is an eigenvector of d/dx, and state its eigenvalue.
Step 2. Is x^2 an eigenvector of d/dx? Explain.
Check your work
Step 1. d/dx(e^(3x)) = 3·e^(3x), which is the eigenvector equation M · v = λ · v with the derivative as M and λ = 3. So e^(3x) is an eigenvector with eigenvalue 3. (In general e^(kx) is an eigenvector with eigenvalue k, which is a large part of why exponentials run through the solutions of differential equations.)
Step 2. No. d/dx(x^2) = 2x, and 2x is not a scalar multiple of x^2 (it is a different function, not just x^2 rescaled). Since differentiating x^2 changes it into something off its own line, x^2 is not an eigenvector of the derivative.
Flashcards
Section titled “Flashcards”Ten cards. Click any card to reveal the answer. Use the Print flashcards button to lay out the full set as one card per page, ready to print or save as a PDF for offline review.
Q. What is the actual definition of a vector?
Anything you can add and scale coherently: add two and get another of the same kind, scale one and get another of the same kind. Arrows and lists are examples, not the definition.
Q. Why are functions vectors?
You can add them pointwise ((f+g)(x) = f(x)+g(x)) and scale them pointwise ((c·f)(x) = c·f(x)), and both stay inside the set of functions. Addition and scaling behave coherently, so functions form a vector space.
Q. What is a basis and dimension for polynomials of degree 3 or less?
Basis {1, x, x^2, x^3}: every such polynomial is a unique combination of these four. The dimension is 4.
Q. How does a polynomial get coordinates?
By fixing a basis. In {1, x, x^2, x^3}, 2x^2 + 5x + 7 is 7·1 + 5·x + 2·x^2 + 0·x^3, so its coordinates are [7, 5, 2, 0]. Adding polynomials becomes adding coordinate vectors.
Q. How is the derivative a matrix?
Differentiation is linear, so it is captured by where it sends the basis. Recording d/dx of each basis polynomial as a column gives the derivative matrix, and differentiating is matrix-vector multiplication.
Q. What is D times [7, 5, 2, 0], and what does it mean?
With the degree-3 derivative matrix, D · [7, 5, 2, 0] = [5, 4, 0, 0], which reads back as 4x + 5, the derivative of 2x^2 + 5x + 7. Calculus carried out by matrix multiplication.
Q. Do coordinates depend on a basis for abstract vectors too?
Yes. A polynomial’s coordinates [7, 5, 2, 0] are relative to the basis {1, x, x^2, x^3}. Choose a different basis for the same space and the same polynomial gets different coordinates, exactly as with arrows.
Q. Which functions are eigenvectors of the derivative?
The exponentials: d/dx(e^(kx)) = k·e^(kx), so e^(kx) is an eigenvector with eigenvalue k. That is much of why exponentials appear throughout the solutions of differential equations.
Q. What from the whole track carries over to any vector space?
Everything: spans, bases, dimension, linear transformations, matrices, the determinant, eigenvectors, change of basis. They are properties of the two rules (add and scale), not of arrows, so they apply to any vector space.
Q. How does abstract vector spaces connect to AI?
ML lives in abstract vector spaces: embeddings (word, sentence, image) are high-dimensional vectors ranked by cosine similarity and compressed by PCA; layers are matrices acting on vectors; function spaces appear in kernels. “Latent space,” “embedding space,” and “function space” all mean a vector space in this sense.