6.1 Eigenvalues and Eigenvectors

1. Verify that $\mathbf{u}$ is an eigenvector for the matrix $A$ and determine the associated eigenvalue. a. $A=\left[\begin{array}{ll}3& 1\\ 1& 3\end{array}\right],\mathbf{u}=\left[\begin{array}{l}1\\ 1\end{array}\right]$

Yes, becomes [4,4]

b. $A=\left[\begin{array}{cc}-7& 10\\ -5& 8\end{array}\right],\mathbf{u}=\left[\begin{array}{l}2\\ 1\end{array}\right]$

Yes, becomes [-4,-2]

c. $A=\left[\begin{array}{ccc}10& 12& -6\\ 8& 11& -4\\ 34& 44& -19\end{array}\right],\mathbf{u}=\left[\begin{array}{l}0\\ 1\\ 2\end{array}\right]$

Yes, becomes [0,3,6]

Copy

Problem 2

a.
Characteristic Polynomial: ( \lambda^2 - 6\lambda + 8 )
Eigenvalues: ( \lambda = 4, 2 )

b.
Characteristic Polynomial: ( \lambda^2 - \lambda - 6 )
Eigenvalues: ( \lambda = 3, -2 )

c.
Characteristic Polynomial: ( (\lambda - 1)^3 )
Eigenvalues: ( \lambda = 1 ) (multiplicity 3)

---

Problem 3

a.
Basis for Eigenspace (λ=2):
[ \left{ \begin{bmatrix}1 \ -1\end{bmatrix} \right} ]

b.
Basis for Eigenspace (λ=1):
[ \left{ \begin{bmatrix}0 \ 3 \ 2\end{bmatrix} \right} ]

c.
Basis for Eigenspace (λ=2):
[ \left{ \begin{bmatrix}1 \ 3 \ 0\end{bmatrix}, \begin{bmatrix}0 \ 1 \ 1\end{bmatrix} \right} ]

---

Problem 4

Characteristic Polynomial: ( (\lambda - 1)^2 )
Eigenvalues: ( \lambda = 1 ) (multiplicity 2)
Basis for Eigenspace:
[ \left{ \begin{bmatrix}1 \ 0\end{bmatrix} \right} ]

---

Problem 5

Characteristic Equation: ( \lambda^2 - \lambda - 6 = 0 )
Verification:
[ A^2 - A - 6I = \begin{bmatrix}0 & 0 \ 0 & 0\end{bmatrix} ]
Thus, ( A ) satisfies its characteristic equation.

---

Problem 6

Formula for Eigenvalues:
[ \lambda = \frac{(a + d) \pm \sqrt{(a - d)^2 + 4bc}}{2} ]

---

Problem 7

Proof:
Let ( A\mathbf{u} = \lambda\mathbf{u} ) and ( B\mathbf{u} = \mu\mathbf{u} ). Then:
[ AB\mathbf{u} = A(B\mathbf{u}) = A(\mu\mathbf{u}) = \mu\lambda\mathbf{u} ]
Hence, ( \mathbf{u} ) is an eigenvector of ( AB ).