Andrew Thangaraj

Aug-Nov 2020

- Vector space \(V\) over a scalar field \(F\)
- \(F\): real field \(\mathbb{R}\) or complex field \(\mathbb{C}\) in this course

- Finite-dimensional vector space
- \(V\) is span of a finite set of vectors

- Basis
- Linearly independent spanning set
- Any two bases have same number of vectors

- Dimension
- Size of basis

*Sum of subsets*: For subsets \(U,W\subseteq V\), the *sum* \(U+W\) is defined as \[U+W = \{u+w:u\in U,w\in W\}\]

Usually, \(U\) and \(W\) are taken to be subspaces. \(U+W\) is the *smallest* subspace containing both \(U\) and \(W\).

*Examples*\(\text{span}((1,0))+\text{span}((0,1))\)

\(\text{span}((1,2))+\text{span}((2,3))\)

\(\text{span}((1,2))+\text{span}((2,4))\)

\(U=\text{span}((1,2,3,4),(2,3,4,5),(1,-1,2,-2))\), \(W=\text{span}((1,1,1,1),(4,1,8,1),(1,0,0,0))\)

\(\text{span}(u_1,\ldots,u_m)+\text{span}(w_1,\ldots,w_m)=\text{span}(u_1,\ldots,u_n,w_1,\ldots,w_m)\)

Find linear dependence in \(u_1,\ldots,u_n,w_1,\ldots,w_m\) to reduce.

*Intersection of subspaces*: If \(U,W\) are subspaces, \(U\cap W\) is a subspace.

*Examples*\(\text{span}((1,0))\cap \text{span}((0,1))=\{0\}\)

\(\text{span}((1,2))\cap \text{span}((2,3))=\{0\}\)

\(\text{span}((1,2))\cap \text{span}((2,4))=\text{span}((1,2))\)

\(U=\text{span}((1,2,3,4),(2,3,4,5),(1,-1,2,-2))\), \(W=\text{span}((1,1,1,1),(4,1,8,1),(1,0,0,0))\)

\(\text{span}(u_1,\ldots,u_n)\cap\text{span}(w_1,\ldots,w_m)=\{v:v=a_1u_1+\cdots+a_nu_n=b_1w_1+\cdots+b_mw_m\}\)

Find linear combinations of \(u_1,\ldots,u_n,w_1,\ldots,w_m\) that result in 0.

*Direct sum of subspaces*: If \(U,W\) are subspaces and \(U\cap W=\{0\}\), \(U+W\) is denoted \(U\oplus W\) and called as direct sum.

Given a subspace \(U\) of a finite-dimensional vector space \(V\), there exists a subspace \(W\) such that \(V=U\oplus W\).

*Proof*- Extend basis of \(U\) to basis of \(V\).
- Define \(W\) as span of new vectors needed in extension.

*Examples*

- \(U=\text{span}((1,2))\)
- \(U=\text{span}((1,2,3))\)
- \(U=\text{span}((1,2,3,4),(2,3,4,5))\)

*Subspace decomposition*: any \(v\in V\) can be written as \(v=u+w\), \(u\in U\), \(w\in W\), in a unique way.

If \(U,W\) are subspaces of a finite dimensional vector space, \[\text{dim}(U+W)=\text{dim} U+\text{dim} W - \text{dim}(U\cap W).\]

*Proof*- Extend basis of \(U\cap W\) to basis of \(U\) and to basis of \(W\).
- Show that vectors above form a basis for \(U+W\).

Numerical examples require methods for the following:

- Find linear dependence between vectors.
- Extend basis.

- Modify a set of vectors to make them look like standard basis without changing the span.
- Modifications are done one step at a time and are reversible.

\(S=((1,2,3,4),(2,3,4,5),(3,4,5,6))\)

- Pivot at first vector, first coordinate (pivot should be nonzero)
- Replace \((2,3,4,5)\) by \((2,3,4,5) - 2(1,2,3,4) = (0,-1,-2,-3)\)
- Replace \((3,4,5,6)\) by \((3,4,5,6) - 3(1,2,3,4) = (0,-2,-4,-6)\)

\(((1,2,3,4),(0,-1,-2,-3),(0,-2,-4,-6))\)

- Pivot at second vector, second coordinate
- Replace \((1,2,3,4)\) by \((1,2,3,4) + 2(0,-1,-2,-3) = (1,0,-1,-2)\)
- Replace \((0,-2,-4,-6)\) by \((0,-2,-4,-6) - 2(0,-1,-2,-3) = (0,0,0,0)\)
- This means that third vector is linearly dependent on first two, and can be dropped

\(((1,0,-1,-2),(0,-1,-2,-3))\) is a linearly independent set with span equal to \(\text{span}(S)\)

- The above form is called “reduced echelon” form
- Extending to basis of \(V\)
- \((1,0,-1,2),(0,1,2,3),(0,0,1,0),(0,0,0,1)\)