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\newcommand{\proj}{\mathrm{proj}} \newcommand{\spn}{\mathrm{span}}

Projection and the Gram-Schmidt Process

The goal this class is to find orthonormal basis for a subspace.

Example:

  • What is an orthonormal basis for the span of {(2,0),(1,1)}\{(2,0),(1,1)\}?

  • How would we do it with 3 vectors?

Definition: Let u,vRnu,v\in \mathbb{R}^n with vv nonzero. Then the projection of u onto v is given by ParseError: KaTeX parse error: Undefined control sequence: \proj at position 1: \̲p̲r̲o̲j̲_v u=\frac{u\cd….

Theorem: Let u,vRnu,v\in \mathbb{R}^n and cc be a nonzero scalar. Then

  • ParseError: KaTeX parse error: Undefined control sequence: \proj at position 1: \̲p̲r̲o̲j̲_v u is in ParseError: KaTeX parse error: Undefined control sequence: \spn at position 1: \̲s̲p̲n̲{v}.

  • ParseError: KaTeX parse error: Undefined control sequence: \proj at position 3: u-\̲p̲r̲o̲j̲_v u is orthogonal to vv

  • if uu is in ParseError: KaTeX parse error: Undefined control sequence: \spn at position 1: \̲s̲p̲n̲{v}, then ParseError: KaTeX parse error: Undefined control sequence: \proj at position 3: u=\̲p̲r̲o̲j̲_v u.

  • ParseError: KaTeX parse error: Undefined control sequence: \proj at position 1: \̲p̲r̲o̲j̲_v u = \proj_cv….

Let SS be a nontrivial subspace with orthogonal basis {v1,,vk}\{v_1,\ldots,v_k\}. Then the projection of uu onto SS is given by ParseError: KaTeX parse error: Undefined control sequence: \proj at position 1: \̲p̲r̲o̲j̲_S u=\sum\sb{i=….

Theorem: Let SS be a nonzero subspace of Rn\mathbb{R}^n with orthogonal basis {v1,,vk}\{v_1,\ldots,v_k\}, and let uu be a vector in Rn\mathbb{R}^n. Then

  • ParseError: KaTeX parse error: Undefined control sequence: \proj at position 1: \̲p̲r̲o̲j̲_S u is in SS.

  • ParseError: KaTeX parse error: Undefined control sequence: \proj at position 3: u-\̲p̲r̲o̲j̲_S u is orthogonal to SS.

  • if uu is in SS, then ParseError: KaTeX parse error: Undefined control sequence: \proj at position 3: u=\̲p̲r̲o̲j̲_S u.

  • ParseError: KaTeX parse error: Undefined control sequence: \proj at position 1: \̲p̲r̲o̲j̲_S u is independent of the choice of orthogonal basis for SS.

Theorem: (The Gram-Schmidt Process) Let SS be a subspace with basis {s1,,sn}\{s_1,\ldots,s_n\}. Define v1,,vkv_1,\ldots,v_k by

  • v1=s1v_1=s_1,

  • ParseError: KaTeX parse error: Undefined control sequence: \proj at position 9: v_2=s_2-\̲p̲r̲o̲j̲\sb{v_1}s_2,

  • ParseError: KaTeX parse error: Undefined control sequence: \proj at position 9: v_3=s_3-\̲p̲r̲o̲j̲\sb{v_1}s_3-\pr…,

  • ...

  • ParseError: KaTeX parse error: Undefined control sequence: \proj at position 9: v_k=s_k-\̲p̲r̲o̲j̲\sb{v_1}s_k-\ld…

Then {v1,,vk}\{v_1,\ldots,v_k\} is an orthogonal basis. To make it orthonormal, just normalize each element.

Example: Find an orthonormal basis for the subspace (1,0,1,1),(0,2,0,3),(3,1,1,5)(1,0,1,1), (0,2,0,3), (-3, -1, 1, 5).