Linear Algebra: Vector Spaces and Transformations

• The following exercises cover concepts in Linear Algebra, including vector spaces, linear transformations, kernels, images, direct sums, and projectors.

Exercise 1

• In a vector space E over field K, for a linear transformation u in L(E), the intersection of the kernel of u and the image of u is {0} if and only if the kernel of u-squared equals the kernel of u.
• Expressed as: Ker(u) ∩ Im(u) = {0} ⇔ Ker(u²) = Ker(u).

Exercise 2

• In a finite-dimensional vector space E, for a linear transformation u in L(E) where rank(u) = rank(u²), then Ker(u) ∩ Im(u) = {0}.

Exercise 3

• Given a vector space E over K with u, v in L(E) such that u + v = Identity(E) and u o v = 0, E is the direct sum of Im(u) and Im(v).
• Expressed as: E = Im(u) ⊕ Im(v).

Exercise 4

• Given a finite-dimensional vector space E over K and u in L(E) such that u² = u, then E is the direct sum of the kernel and image of u.
• E = Ker(u) ⊕ Im(u).

Exercise 5

• Considers a finite-dimensional vector space E over K, with F and G being subspaces such that E is the direct sum of F and G (E = F ⊕ G).
• Let p be the projection onto F parallel to G.
• p is a linear transformation in L(E).
• p² = p, meaning p is a projection.
• Image of p equals F, and the kernel of p equals G: Im(p) = F and Ker(p) = G.

Exercise 6

• Given vector space E over a field K, and u in L(E) is a projector.
• The transformation Identity(E) - u is also a projector.
• Need to determine/specify its image and kernel.

Exercise 7

• Given finite-dimensional vector space E over field K, and u in L(E) with rank(u) = 1.
• There exists λ in K such that u² = λu.

Exercise 8

• Given finite-dimensional vector space E over field K, and u in L(E) such that rank(u) = 1, the kernel of u is a subset of the kernel of u-squared.
• Expressed as: Ker(u) ⊂ Ker(u²).

Exercise 9

• Given finite-dimensional vector space E over field K, and u, v in L(E) such that u o v = 0.
• rank(u + v) = rank(u) + rank(v) if and only if the intersection of the image of u and the image of v is {0}.
• Expressed as: rank(u + v) = rank(u) + rank(v) ⇔ Im(u) ∩ Im(v) = {0}.

Theodor Schwann

• Was a German physiologist from 1810-1882
• He laid the foundation of cell theory which identifies the cell as the basic unit of animal structure
• Schwann defined the three structual parts of a cell as the wall, nucleus, and fluid content
• He recognized that these parts were present in plant and animal cells
• He published paper in 1839 called Microscopic Investigation on the Accordance in the Structure and Growth of Animals and Plants
• He famously observed that "all living things are composed of cells and cell products."
• The Schwann cell, which forms the myelin surrounding the axons of peripheral nerves, was named after him.