Section 5.1 Eigenvectors and Eigenvalues

Eigenvectors and Eigenvalues Useful throughout pure and applied mathematics. Used to study difference equations and continuous dynamical systems. Provide critical information in engineering design Arise naturally in such fields as physics and chemistry. Used in statistics to analyze multicollinearity

Example: Let Consider Av for some v

Definition: The Eigenvector of A nxn is a nonzero vector x such that Ax= λ x for some scalar λ. λ is called an Eigenvalue of A

Statistics (multicollinearity) Where y is the dependent response vector and the x’s are the independent explanatory vectors The β ’s are least squares regression coefficients ε i are errors We desire linear independence between x vectors Can use Eigen analysis to determine

From Definition: Ax = λ x = λ Ix Ax – λ Ix = 0 (A – λ I)x = 0 Observations: 1. λ is an eigenvalue of A iff (A – λ I)x= 0 has non-trivial solutions A – λ I is not invertible IMT all false 2. The set {x ε R n : (A – λ I)x= 0} is the nullspace of (A – λ I)x= 0, A a subspace of R n 3. The set of all solutions is called the eigenspace of A corresponding to λ

Example Show that 2 is an eigenvalue of and find the corresponding eigenvectors.

Comments: Warning: The method just used (row reduction) to find eigenvectors cannot be used to find eigenvalues. Note: The set of all solutions to (A- λ I)x = 0 is called the eigenspace of A corresponding to λ.

Example Let An eigenvalue of A is λ =3. Find a basis for the corresponding eigenspace.

Theorem The eigenvalues of a triangular matrix are the entries on its main diagonal.

Proof of 3x3 case Let So (A- λ I) =

Proof of 3x3 case By definition λ is an eigenvalue iff (A- λ I)x= 0 has non-trivial solutions so a free variable must exist. This occurs when a 11 = λ or a 22 = λ or a 33 = λ

Example Consider the lower triangular matrix below. λ = 4 or 0 or -3

Addtion to IMT A nxn is invertible iff s. The number 0 is not an eigenvalue t. det A≠ 0 (not sure why author waits until not to add this)

Theorem If eigenvectors have distinct eigenvalues then the eigenvectors are linearly independent This can be proven by the IMT

Section 5.2 The Characteristic Equation

Finding Eigenvalues 1. We know (A- λ I)x= 0 must have non- trivial solutions and x is non-zero. That is free variables exist. 2. So (A- λ I) is not invertible by the IMT 3. Therefore det(A- λ I)= 0 by IMT

Characteristic Equation det(A- λ I)= 0 Solve to find eigenvalues Note: det(A- λ I) is the characteristic polynomial.

Previous Example: Let find eigenvalues

Example Find Eigenvalues

Example Find characteristic polynomial and eigenvalues

Example a. Find the characteristic polynomial b. Find all eigenvalues c. Find multiplicity of each eigenvalue

Recap a. λ is an eigenvalue of A if (A- λ I)x= 0 has non-trivial solutions (free variables exist). b. Eigenvectors (eigenspace )are found by row reducing (A- λ I)x= 0. c. Eigenvalues are found by solving det(A- λ I)= 0.

Section 5.3 Diagonalization

Diagonalization The goal here is to develop a useful factorization A=PDP -1, when A is nxn. We can use this to compute A k quickly for large k. The matrix D is a diagonal matrix (i.e. entries off the main diagonal are all zeros).

Example Find a formula for A k given A=PDP -1 &

Diagonalizable Matrix “A” is diagonalizable if A=PDP -1 where P is invertible and D is a diagonal matrix. Note: AP=PD

When is a matrix diagonalizable? Let’s examine eigenvalues and eigenvectors of A

The Diagonalization Theorem If Anxn & has n linearly independent eigenvectors. Then 1. A=PDP Columns of P are eigenvectors 3. Diagonals of D are eigenvalues.

Example Diagonalize We need to find P & D

Theorem If Anxn has n distinct eigenvalues then A is diagonalizable.

Example Diagonalize

Example