Traces and Eigenvalues $3\times 3 $ matrix

Question

I understand how to find the trace of a $2\times 2 $ matrix but I am confused with how to find the characteristic polynomial of a general $3\times 3 $ matrix with arbitrary constants.

Answer 1

There are three numbers, call $\sigma_1$ the trace, then call $\sigma_3$ the determinant. The last one, $\sigma_2,$ is the sum of the three two by two submatrices with both diagonal elements on the main diagonal of the original matrix, so $$ \sigma_2 = (a_1 b_2 - a_2 b_1) + (a_1 c_3 - a_3 c_1) + (b_2 c_3 - b_3 c_2) $$

Then the characteristic polynomial is $$ \lambda^3 - \sigma_1 \lambda^2 + \sigma_2 \lambda - \sigma_3. $$ Watch out for the alternating signs, I get that wrong sometimes.

Answer 2

The characteristic polynomial of any matrix, A, is the determinant of the matrix $A- \lambda I$ where I is the identity matrix. In particular, if $A= \begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{bmatrix}$ then the characteristic polynomial is $\left|\begin{array}( a_{11}- \lambda & a_{12} & a_{13} \\ a_{21} & a_{22}- \lambda & a_{23} \\ a_{31} & a_{32} & a_{33}- \lambda\end{array}\right|$.

Now, do you know how to evaluate a 3 by 3 determinant? The simplest way is to "expand" on a row or column. For example, expanding this on the first row gives $(a_{11}- \lambda)\left|\begin{array}( a_{22}- \lambda & a_{23} \\ a_{32} & a_{33}- \lambda \end{array}\right|$$- a_{12}\left|\begin{array}( a_{21} & a_{23} \\ a_{31} & a_{33}- \lambda \end{array}\right|$$+ a_{13}\left|\begin{array}(a_{21} & a_{22}- \lambda \\ a_{31} & a_{32}\end{array}\right|$

Answer 3

A calculation yields the following explicit formula for $\chi_A(t)=\det (tI-A)$: $$ \chi_A(t)= t^3 - t^2(a_1 + b_2 + c_3) + t(a_1b_2 + a_1c_3 - a_2b_1 - a_3c_1 + b_2c_3 - b_3c_2) - a_1b_2c_3 + a_1b_3c_2 + a_2b_1c_3 - a_2b_3c_1 - a_3b_1c_2 + a_3b_2c_1. $$ One can use the rule of Sarrus for a $3\times 3$-determinant, or just a CAS.