TR3.5

2C Proof Presentation

Huxley Marvit 2021-10-10 Sun 20:36

Table of Contents

#ret #hw

1 Proof Prez

The problem:

Suppose \(U_1, ... , U_m\) are finite-dimensional subspaces of \(V\). Prove that \(U_1+ ... + U_m\) is finite-dimensional \[\dim(U_1+ ... + U_m) \leq \dim U_1 + ... + \dim U_m \]

  • 1.39:
    • sum of subspaces is the smallest containing subspace.
      • Suppose \(U_1, ... , U_m\) are subspaces of \(V\). Then \(U_1+ ... + U_m\) is the smallest subspace of V containing \(U_1, ... , U_m\).
  • 2.26
    • Finite-dimensional subspaces
      • every subspace of a finite-dimensional vector space is finite-dimensional
  • 2.43
    • dimension of a sum
      • if \(U_1\) and \(U_2\) are subspaces of a finite-dimensional vector space, then \[ \dim(U_1+U_2)=\dim U_1+\dim U_2 - dim(U_1 \cap U_2) \]
  • 1.8
    • definiton list, length
    • supposed n is a non-negative integer
  • 2.36
    • dimension
      • dimension of a finite dimensional vector space is the lenght of any basis of the vector space
  1. proving finite dimensional

    By 1.39, we know that the sum of subspaces in \(V\) is a subspace in \(V\). By 2.26, we know that every subspace of a finite dimensional vector space is finite-dimensional V is finite dimensional

    therefore, sum of subspaces in \(V\) is finite-dimensional

  2. proving dim

    by 2.36, we know that the dimension is the lenght of the basis by 1.8, we know that a lenght cannot be negative thus, dim U1 + dim U2 - dim (U1 intsct U2) will always be <= dim U1 + dim U2 – can't subtract and a positive number and get a larger number.

    by 2.43, we know that dim(U1+U2) = dim U1 + dim U2 - dim (U1 intsct U2) therefore, dim(U1+U2) will always be <= dim U1 + dim U2

    generalize to list: by 1.39, we know that the sum of subspaces in V is a subspace of V so, let U1 = u1, and U2 = u2 + … + um thus, dim(U1 + U2) = dim(u1 … um)

    therefore, dim(u1 … um) <= dim u1 … um

1.1 Formalizing