MATH 323 Lecture 24

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Theorem 5.6.1 (Gram-Schmidt Process)

Let be a basis for . Let and define recursively by

Where is the projection of onto .

The set is an orthonormal basis for .


From now on,

Proof by Induction

and is an orthonormal basis for

Suppose are constructed so they are an orthonormal basis for and .

because are linearly independent.

, where

Therefore is an orthonormal set of vectors in , and is a basis for subspace of dimension .

Thus the theorem holds by induction.



Find an orthonormal basis for the range of the following matrix:

Column space is defined by

Therefore, is an orthonormal basis for the range of .

Theorem 5.6.2 (Gram-Schmidt QR Factor)

If is an matrix of rank , then can be factored into a product , where is an matrix with orthonormal column vectors and is an upper triangular matrix whose diagonal entries are all positive.

In the Gram-Schmidt Orthogonalization Process, represents the projection vectors.

We use the defined values above to form an upper triangular matrix

such that , where .


From previous example,

Polynomial Space Example

Consider , the subspace consisting of quadratic polynomials, with an inner product:

, where

, , and (represented by ) form a basis for , but they are not orthonormal w.r.t. the inner product definition.

Find an orthonormal basis for .

Eigenvalues and Eigenvectors

Let be an matrix.

A scalar is an eigenvalue of if there is a nonzero vector such that

is said to be an eigenvector belonging to .

All of the following are equivalent: is an eigenvalue iff

  • has nonzero solution
  • This is important since it gives the characteristic equation
  • is singular




Therefore, is the eigenvalue, and is an eigenvector belonging to .

If is an eigenvector belonging to , then is also an eigenvector belonging to (for nonzero ):

Characteristic Equation

is called the characteristic polynomial

is called the characteristic equation

Either way, the form is

Example 1

Characteristic equation is .

Solutions are .

Eigenvectors can be obtained by solving .