Easier way to find eigenvalues of Matrices?

Question

I am trying to find eigenvalues for this matrix,
A = $\begin{bmatrix} 3 & 2 & -3 \\ -3 & -4 & 9 \\ -1 & -2 & 5 \\ \end{bmatrix}$

I find the characteristic equation here:
$(\lambda I - A) = \begin{bmatrix} \lambda - 3 & -2 & 3 \\ 3 & \lambda + 4 & -9 \\ 1 & 2 & \lambda - 5 \\ \end{bmatrix}$

The difficult part I am having is finding the determinant of the characteristic equation, mainly that it becomes insanely difficult for me to keep track of the factoring to get the eigenvalues...here I use rule of Sarrus to try to calculate the eigenvalues. $\begin{vmatrix} \lambda I - A \\ \end{vmatrix} = \begin{bmatrix} \lambda - 3 & -2 & 3 & \lambda -3 & -2 \\ 3 & \lambda + 4 & -9 & 3 & \lambda + 4 \\ 1 & 2 & \lambda - 5 & 1 & 2 \\ \end{bmatrix}$
= $(1)(\lambda + 4)(3)$
$+ (2)(-9)(\lambda - 3)$
$+ (\lambda - 5)(3)(-2)$
$- (\lambda - 3)(\lambda + 4)(\lambda - 5)$
$- (-2)(-9)(1)$
$- (3)(3)(2)$

= $(3)(\lambda + 4) + (-18)(\lambda - 3) + (-6)(\lambda -5) - (\lambda - 3)(\lambda + 4)(\lambda - 5) - 18 - 18$
= $(3)(\lambda + 4) + (-18)(\lambda - 3) + (-6)(\lambda -5) - (\lambda - 3)(\lambda + 4)(\lambda - 5) - 36$
= $((\lambda + 4)(3) -(\lambda -3)(\lambda -5)) + (-18)(\lambda -3) + (-6)(\lambda - 5) - 36$
= $(\lambda - 3)((\lambda + 4)(3) -(\lambda - 5)) -18 + (-6)(\lambda - 5) - 36$
= $(\lambda - 5)(\lambda -3)((\lambda + 4)(3) - 1)) - 18 + (-6)(1) - 36$
= $(\lambda - 5)(\lambda -3)((\lambda + 4)(3) - 1)) - 60$
= $(\lambda - 5)(\lambda - 3)(3\lambda + 12 -1 - 60)$
= $(\lambda - 5)(\lambda - 3)(3\lambda - 49)$

I end up with 5, 3, and 16.3 as the eigenvalues (16.3 seems off). Obviously that was a ridiculous amount of simplification I had to do just to get eigenvalues and when I feel like I messed up (like here), it is pretty impossible to check my work in an effective manner. The amount of time it takes for me to calculate eigenvalues is unacceptable for my upcoming final exam. Do you guys have any tips or tricks that makes this process easier?

Answer 1

Here is a simple computation: \begin{align} \det(\lambda I - A)& = \begin{vmatrix} \lambda - 3 & -2 & 3 \\ 3 & \lambda + 4 & -9 \\ 1 & 2 & \lambda - 5 \\ \end{vmatrix}=\begin{vmatrix} \lambda - 2 & 0 & \lambda - 2 \\ 3 & \lambda + 4 & -9 \\ 1 & 2 & \lambda - 5 \\ \end{vmatrix}\\& =(\lambda - 2 )\begin{vmatrix} 1 & 0 & 1 \\ 3 & \lambda + 4 & -9 \\ 1 & 2 & \lambda - 5 \\ \end{vmatrix}=(\lambda - 2 )\biggl(\begin{vmatrix} \lambda + 4 & -9 \\ 2 & \lambda - 5 \\ \end{vmatrix}+\begin{vmatrix} 3 & \lambda + 4 \\ 1 & 2 \\ \end{vmatrix}\biggl)\\ &=(\lambda - 2 )\Bigl(\bigl( (\lambda + 4)(\lambda - 5)+18\bigr) + (6 -\lambda - 4) \Bigl)=(\lambda - 2 )(\lambda^2-2\lambda)\\ &=\lambda(\lambda - 2)^2. \end{align}

Answer 2

For 2 by 2 and 3 by 3 there are worthwhile recipes for the characteristic polynomial. For 2 by 2, it is $\lambda^2 - T \lambda + D,$ where $T$ is the trace and $D$ is the determinant.

more in a few minutes.

For 3 by 3, it is $$ \lambda^3 - \sigma_1 \lambda^2 + \sigma_2 \lambda - \sigma_3 $$ Here $\sigma_1$ is the trace and $\sigma_3$ is the determinant. The middle one, $\sigma_2$ is the sum of the three two by two mini-determinants with both diagonal entries on the main diagonal. With your original matrix, we have $$ \sigma_2 = \left| \begin{array}{rr} 3 & 2 \\ -3 & -4 \\ \end{array} \right| + \left| \begin{array}{rr} 3 & -3 \\ -1 & 5 \\ \end{array} \right| + \left| \begin{array}{rr} -4 & 9 \\ -2 & 5 \\ \end{array} \right| = -6 + 12 -2 = 4 $$

We have the familiar trace $\sigma_1 = 4,$ and the determinant $\sigma_3 = 0$ by that Rule of Sarrus. $$ \lambda^3 - 4 \lambda^2 + 4 \lambda - 0 = \lambda (\lambda- 2 )^2 $$

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Note that the same method, done carefully, works in any dimension. In dimension 4, we get $$ \lambda^4 - \sigma_1 \lambda^3 + \sigma_2 \lambda^2 - \sigma_3 \lambda + \sigma_4. $$ Once again $\sigma_1$ is the trace and $\sigma_4$ the determinant. This time we have two middle terms. Now $\sigma_2$ is the sum of (determinants of) six little 2 by 2 matrices along the diagonal, while $\sigma_3$ is the sum of (determinants of) four little 3 by 3 matrices along the diagonal.

Answer 3

I would do what you did: to apply the rule of Sarrus. But the polynomial that you got cannot be correct. First of all, it is not monic. On the other hand, if you sum all entries of each row of the matrix, you get $2$. But that means that $2$ is an eigenvalue and that $(1,1,1)$ is an eigenvector of $A$. However, $2$ is not a root of your polynomial.