Answer :
Given:
[tex]A=\begin{bmatrix}{4} & {6} & {10} \\ {3} & {10} & {13} \\ {-2} & {-6} & {-8}\end{bmatrix}[/tex]First we need to find the eigenvalues of A. Recall that they are the solutions of the equation det(λI - A) = 0:
[tex]det\lparenλ\begin{bmatrix}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1}\end{bmatrix}-\begin{bmatrix}{4} & {6} & {10} \\ {3} & {10} & {13} \\ {-2} & {-6} & {-8}\end{bmatrix})=0[/tex]which becomes
[tex]det\lparen\begin{bmatrix}{λ}-4 & {-6} & {-10} \\ {-3} & {λ-10} & {-13} \\ {2} & {6} & {λ+8}\end{bmatrix})=0[/tex]Calculate this determinant:
[tex]\begin{gathered} (λ-4)(λ-10)(λ+8)+(-3)(6)(-10)+(2)(-6)(-13) \\ -(-10)(λ-10)(2)-(-6)(-3)(λ+8)-(-13)(6)(λ-4)=0 \end{gathered}[/tex]Simplify:
[tex]λ^3-6λ^2+8λ=0[/tex]Then, factor:
[tex]λ(λ-2)(λ-4)=0[/tex]Separate the solutions:
[tex]\begin{gathered} λ=0\text{ or} \\ λ-2=0 \\ λ-2+2=0+2 \\ λ=2\text{ or} \\ λ-4=0 \\ λ-4+4=0+4 \\ λ=4 \end{gathered}[/tex]Now that we have found the eigenvalues for A , we can compute the eigenvectors:
For λ = 0
[tex]\begin{bmatrix}{-1} & & \\ {-1} & & \\ {1} & & \end{bmatrix}[/tex]For λ = 2
[tex]\begin{bmatrix}{} & {1} & {} \\ {} & {-2} & {} \\ {} & {1} & {}\end{bmatrix}[/tex]For λ = 4
[tex]\begin{bmatrix}{} & {-3} & {} \\ {} & {-5} & {} \\ {} & {3} & {}\end{bmatrix}[/tex]Answer:
The eigenvalues are:
[tex]\begin{gathered} λ=0 \\ λ=2 \\ λ=4 \end{gathered}[/tex]And the eigenvectors are:
[tex]\begin{bmatrix}{} & -{1} & {} \\ {} & {-1} & {} \\ {} & {1} & {}\end{bmatrix},\begin{bmatrix}{} & {1} & {} \\ {} & {-2} & {} \\ {} & {1} & {}\end{bmatrix},\begin{bmatrix}{} & {-3} & {} \\ {} & {-5} & {} \\ {} & {3} & {}\end{bmatrix}[/tex]