Sums, Direct Sums and Gaussian Elimination

Aug-Nov 2020

Recap

• 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 subspaces

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

1. $\text{span}((1,0))+\text{span}((0,1))$

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

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

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

Intersection of subspaces: If $U,W$ are subspaces, $U\cap W$ is a subspace.

• Examples

1. $\text{span}((1,0))\cap \text{span}((0,1))=\{0\}$

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

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

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

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

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

Direct sum of subspaces

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

1. $U=\text{span}((1,2))$
2. $U=\text{span}((1,2,3))$
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.

Dimension of sum of two subspaces

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.

Gaussian Elimination: Implementing Linear Dependence Lemma

• 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 $V$
• $(1,0,-1,2),(0,1,2,3),(0,0,1,0),(0,0,0,1)$