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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.
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:
, , 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 ):
is called the characteristic polynomial
is called the characteristic equation
Either way, the form is
Characteristic equation is .
Solutions are .
Eigenvectors can be obtained by solving .