Finding the Characteristic Equation

Question

For the following matrix I need to find

$$\begin{bmatrix}-3 & 2 &1 \\3 & -4 & -3 \\-8 & 8 & 6 \end{bmatrix}$$

a. Characteristic Polynomial of $A$

b. Eigen Values

c. Eigen Vectors

d. is $A$ diagonlizable?

But In order to find a-d, first I need to find the characteristic polynomial Heres, where I am stuck, =>$\text{det}(\lambda I-A) = 0$ I get the following, $$\begin{bmatrix}\lambda+3 & -2 & -1 \\-3 & \lambda+4 & 3 \\8 & -8 & \lambda-6 \end{bmatrix}$$

Solving for it I am stuck on the factoring

Solving the determinant by row 1 I get,

$\lambda+3((\lambda +4)(\lambda -6)+24)+2(-3(\lambda -6)-24)-1(24-8(\lambda +4)) = 0$

After this I am unable to factor, thereby I am not able to solve the entire question please help

Answer 1

$(λ+3)((λ +4)(λ -6)+24)+2(-3(λ -6)-24)-1(24-8(λ +4)) $

$=(λ^3+λ^2-30λ-72)+24(λ+3) +2(-3λ-6)-(-8-8λ)$

$=λ^3+λ^2-30λ-72+24λ+72-6λ-12+8+8λ$

$=λ^3+λ^2-4λ-4$

$=(λ+1)(λ^2-4)$Since we can spot $λ=-1$ as a root

$=(λ+1)(λ-2)(λ+2)$

Answer 2

You can use $\begin{vmatrix}\lambda+3&-2&-2\\-3&\lambda+4&3\\8&-8&\lambda-6\end{vmatrix}=\begin{vmatrix}\lambda+1&-2&-1\\\lambda+1&\lambda+4&3\\0&-8&\lambda-6\end{vmatrix}=(\lambda+1)\begin{vmatrix}1&-2&-1\\1&\lambda+4&3\\0&-8&\lambda-6\end{vmatrix}$

$\hspace{.7 in}=(\lambda+1)\begin{vmatrix}1&-2&-1\\0&\lambda+6&4\\0&-8&\lambda-6\end{vmatrix}=(\lambda+1)\begin{vmatrix}\lambda+6&4\\-8&\lambda-6\end{vmatrix}$,

(adding C2 to C1, factoring out $\lambda+1$, and subtracting R1 from R2)

Answer 3

First of all, note that $\lambda+3$ should be in parentheses in your expression for the determinant.

Next, observe that, for example, $$(\lambda+4)(\lambda-6)+24=\lambda^2-2\lambda-24+24=\lambda^2-2\lambda.$$ Similarly, you should simplify the quantities within each other pair of parentheses. Next, expand all the products and gather like terms. Then you can start factoring.

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Consider the simpler matrix $$B=\begin{bmatrix}-4 & -3\\8 & 6\end{bmatrix}.$$ Readily, $$\det(\lambda I_2-B)=(\lambda+4)(\lambda-6)+24,$$ but if we leave it in that form, it isn't at all clear how we can factor it! We have to expand the product $$(\lambda+4)(\lambda-6)=\lambda^2-2\lambda-24$$ to get that $$\det(B-\lambda I_2)=\lambda^2-2\lambda=\lambda(\lambda-2).$$