We find the eigenvectors associated with each of the eigenvalues • Case 1: λ = 4 – We must find vectors x which satisfy (A −λI)x= 0. Then λ 0 ∈ C is an eigenvalue of the problem-if and only if F (λ 0) = 0. Both Theorems 1.1 and 1.2 describe the situation that a nontrivial solution branch bifurcates from a trivial solution curve. But all other vectors are combinations of the two eigenvectors. A vector x perpendicular to the plane has Px = 0, so this is an eigenvector with eigenvalue λ = 0. :5/ . Other vectors do change direction. Since there are three distinct eigenvalues, they have algebraic and geometric multiplicity one, so the block diagonalization theorem applies to A. Eigenvectors and eigenvalues λ ∈ C is an eigenvalue of A ∈ Cn×n if X(λ) = det(λI −A) = 0 equivalent to: • there exists nonzero v ∈ Cn s.t. If λ 0 ∈ r(L) has the above properties, then one says that 1/λ 0 is a simple eigenvalue of L. Therefore Theorem 1.2 is usually known as the theorem of bifurcation from a simple eigenvalue; it provides a much better description of the local bifurcation branch. Proof. Similarly, the eigenvectors with eigenvalue λ = 8 are solutions of Av= 8v, so (A−8I)v= 0 =⇒ −4 6 2 −3 x y = 0 0 =⇒ 2x−3y = 0 =⇒ x = 3y/2 and every eigenvector with eigenvalue λ = 8 must have the form v= 3y/2 y = y 3/2 1 , y 6= 0 . The first column of A is the combination x1 C . T ( v ) = λ v. where λ is a scalar in the field F, known as the eigenvalue, characteristic value, or characteristic root associated with the eigenvector v. Let’s see how the equation works for the first case we saw where we scaled a square by a factor of 2 along y axis where the red vector and green vector were the eigenvectors. Determine a fundamental set (i.e., linearly independent set) of solutions for y⃗ ′=Ay⃗ , where the fundamental set consists entirely of real solutions. detQ(A,λ)has degree less than or equal to mnand degQ(A,λ)

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