Section 3.3 Diagonalization and Eigenvalues
3.3A
How do we find the matrix [latex]A^100[/latex]? The application for this is that sometimes we need to apply the same process a couple of times and see what it would become.
Example: Consider the evolution of the population of a species of birds. Because the number of males and females are nearly equal, we count only females. We assume that each female remains a juvenile for one year and then becomes an adult and that only adults have offspring. We make three assumptions about reproduction and survival rates:
1. The number of juvenile females hatched in any year is twice the number of adult females alive the year before (we say the reproduction rate is 2).
2. Half of the adult females in any year survive to the next year (the adult survival rate is 1/2)
3. One-quarter of the juvenile females in any year survive into adulthood
(the juvenile survival rate is 1/4.)
If there were 100 adult females and 40 juvenile females alive initially,
compute the population of females [latex]k[/latex] years later.
Definition: An eigenvector of an [latex]n \times n[/latex] matrix [latex]A[/latex] is a nonzero vector [latex]\vec{x}[/latex] such that for some scalar [latex]\lambda[/latex], [latex]A\vec{x} = \lambda\vec{x}[/latex]. A scalar [latex]\lambda[/latex] is called an eigenvalue of [latex]A[/latex] if there is a nontrivial solution [latex]\vec{x}[/latex] of [latex]A\vec{x} = \lambda\vec{x}[/latex]; such an [latex]\vec{x}[/latex] is called an eigenvector corresponding to [latex]\lambda[/latex].
Remark: [latex]\lambda[/latex] is an eigenvalue of an matrix [latex]A[/latex] if and only if the equation [latex](A - I\lambda)\vec{x} = 0[/latex] has a nontrivial solution
Example 1: Show that [latex]-2[/latex] is an eigenvalue of matrix [latex]\begin{bmatrix}-1 & 2\\3 & 4\end{bmatrix}[/latex], and find the corresponding eigenvectors.
Exercise 1: Show that [latex]3[/latex] is an eigenvalue of matrix [latex]\begin{bmatrix}2 & 3\\-2 & 9\end{bmatrix}[/latex], and find the corresponding eigenvectors.
Example 2: Show that [latex]-3[/latex] is an eigenvalue of matrix [latex]\begin{bmatrix}1 & 1 & 2\\3 & 0 & 6\\-2 & 2 & 1\end{bmatrix}[/latex] and find eigenvectors corresponding to the eigenvalue [latex]-3[/latex].
Exercise 2: Show that [latex]4[/latex] is an eigenvalue of matrix [latex]\begin{bmatrix}6 & -4 & 6\\1 & 2 & 3\\-3 & 6 & -5\end{bmatrix}[/latex] and find eigenvectors corresponding to the eigenvalue [latex]4[/latex].
Question: How do we find the eigenvalues?
Theorem: The eigenvalues of a triangular matrix are the entries on its main diagonal.
Proof:
Remark: Unfortunately, we cannot reduce a non-triangular matrix to echelon or triangular matrix to find the eigenvalue of a matrix [latex]A[/latex]. [latex]-5[/latex] is an eigenvalue of matrix [latex]\begin{bmatrix}-4 & -3\\4 & -17\end{bmatrix}[/latex] but its’ echelon form is [latex]\begin{bmatrix}-4 & -3\\0 & -20\end{bmatrix}[/latex] which has eigenvalues [latex]-4[/latex]