Please check my work in finding eigenvalues for the following problem. I am working out of the textbook Advanced Engineering Mathematics by Erwin Kreyszig, 1988, John Wiley & Sons.
For reference the following identity is given because some textbooks reverse the formula having $\lambda$ subtract the diagonal elements instead of subtracting $\lambda$ from the diagonal elements:
$$ det(A – \lambda I) =
\begin{vmatrix}
13-\lambda & 0 & -15 \\
-3 & 4-\lambda & 9 \\
5 & 0 & -7-\lambda \\
\end{vmatrix} = 0
$$
Taking the center column we have:
$$
(4-\lambda)
\begin{vmatrix}
13-\lambda & -15 \\
5 & -7-\lambda \\
\end{vmatrix} \\
= (4-\lambda)[(13-\lambda)(-7-\lambda) + 5(15)] = 0 \\
= (4-\lambda)(\lambda^2 – 6\lambda – 91 + 75) = 0 \\
= (4-\lambda)(\lambda^2 – 6\lambda -21) = 0 \\
= -\lambda^3 + 10\lambda^2 – 24\lambda + 21\lambda – 84 = 0 \\
= \lambda^3 – 10\lambda^2 + 3\lambda + 84 = 0 \\
$$
Using an online calculator the characteristic equation factors into:
$$
\lambda^3 – 10\lambda^2 + 3\lambda + 84 = 0 \\
(\lambda – 4)(\lambda^2 – 6\lambda – 21) = 0 \\
$$
But answer in text is
$$
\begin{align*}
\lambda_1 = 8 \qquad \lambda_2 = 4 \qquad \lambda_3 = -2 \\
\end{align*}
$$
Question: Although I can get $\lambda = 4$ out of the factored equation there is no way to get the other two eigenvalues. I suspect my algebra. Where did I go wrong?
Best Answer
Your error is in the step $$(4-\lambda)(\lambda^2 - 6\lambda - 91 + 75) = 0 \\ = (4-\lambda)(\lambda^2 - 6\lambda \color{red}{-21}) = 0$$ It should be $\color{blue}{-16}$ and then it factors well to give you the required answer