Better Method for factoring involving Characteristic Polynomial

characteristic polynomialfactoringlinear algebra

I am in a linear algebra class right now, and I am reviewing for diagonalization. With my final coming up I am looking to cut down time spent on finding the eigenvalues of $A$ said matrix $A$.

An example we have done in class is as follows:

Let $A= \begin{bmatrix}3&1\\1&3\end{bmatrix}$ This is a simple example of a problem, but my question is:

When solving I get: $\det(A-I_2)$ I get $(3-λ)(3-λ) – 1$.

My method of solving for $λ$'s here is as follows:

FOIL $(3-λ)(3-λ) – 1$

= $\lambda^2 -3λ – 3λ + 9 – 1$

= $λ^2 -6λ + 8$

Here I refactor and get: $(λ – 4)(λ – 2)$
Solving for lambda I get $λ= 2, 4$.

My professor in class seems to have sipped the entire foil process and just subtracted one from the first term and added one to the second and produced the same $(λ – 4)(λ – 2)$.

This may not seem like a huge deal on a problem like this, but when dealing with cubic polynomials if I don't have to foil 3 out it will

  1. Save me time

  2. Provide less chance for mistakes.

My question at its core is,

Is there a better way to solve for $\lambda$ given: $(3-λ)(3-λ) – 1$ besides foiling and then refactoring?

Best Answer

In general for a matrix of order $n\times n$ with $n\geqslant 3$, find the characteristic polynomial is based on factoring and sometimes spending time on brute force along with synthetic division or using root rational theorem.

However we know special cases, for example.

  • On order $2\times 2$ with $A=\begin{bmatrix} a & b\\ c & d\end{bmatrix}$ then the characteristic polynomial is given by $p(\lambda)=\lambda^2 -{\rm Tr}(A)\lambda+\det(A)$ as pointed out Ted Shifrin on comment.
  • Upper or lower triangular matrix for example $A=\begin{bmatrix} a& b & c\\ 0 & d & e\\ 0 & 0 & f\end{bmatrix}$ then the characteristic polynomial is given by $p(\lambda)=(a-\lambda)(d-\lambda)(f-\lambda)$.
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