How to find the characteristic polynomial of a $3 \times 3$ matrix?

Question

What is the best strategy to find the characteristic polynomial of a 3x3 matrix?

For example I got this matrix:

$A=\begin{bmatrix}6 & -2 & -1 \\ -2 & 6 & -1 \\ -1 & -1 & 5 \end{bmatrix}$

and I know the characteristic equation of this matrix should be: $ -\lambda^3 + 17\lambda^2 -90\lambda + 144 $ and that I can get it from solving $(6-\lambda)(6-\lambda)(5-\lambda)+something$ however the formula without $something$ solves to $-\lambda^3 + 17 \lambda^2 - 96\lambda + 180$. What should this $something$ be?

Alternatively, this website says this formula exists:

$−λ^3+tr(A)λ^2+(tr(A)^2−tr(A^2))λ+det(A)$

however, I find:

$A^2=\begin{bmatrix}41 & -23 &-9 \\ -23 & 41 & -9 \\ -9 & -9 & 27 \end{bmatrix}$

and thus $tr(A)^2−tr(A^2)=17^2-109=180$ which results in the equation $−λ^3+17λ^2+180λ+144$ which is not correct. Did I do something wrong here? Because if this formula does work it is very usefull

Answer 1

You got the matrix $A$, and you know that its characteristic polynomial should be $ -\lambda^3 + 17\lambda^2 -90\lambda + 144 ,$ but when you apply the formula listed at that web site, you get $−λ^3+17λ^2+\color{red}{180}λ+144$, which is not correct. You wrote:

Did I do something wrong here? Because if this formula does work it is very useful

You did not do anything wrong, but the formula listed on that web site left out a factor of $\frac12$, which should multiply the coefficient $tr(A)^2-tr(A^2)$ of $\lambda$. Of note, that web site seems to calculate the characteristic polynomial correctly when the matrix components are entered. Correct formulas for the characteristic polynomial of a $3\times3$ matrix, including $\frac12[tr(A)^2-tr(A^2)],$ are given on Mathworld.

Answer 2

Since asking this question I did some calculating. I like to share it. Hopefully it makes some sense...

For some reason finding the 2x2 minors never felt intuitive to me. I understand that it works but I don't understand what/why I do it.

For me row reducing feel more understandable. For example the $det(A)=ad-bc$:

$A=\begin{bmatrix}a & b \\ c & d \end{bmatrix}\sim\begin{bmatrix}a & b \\ 0 & d-\frac{c}{a}b \end{bmatrix}, det(A)=a(d-\frac{c}{a}b) =ad-bc $

and with the same method:

$B=\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\sim \begin{bmatrix}a & b & c \\ 0 & e-\frac{d}{a}b & f-\frac{d}{a}c \\ 0 & h-\frac{g}{a}b & i-\frac{g}{a}c \end{bmatrix}\sim \begin{bmatrix}a & b & c \\ 0 & e-\frac{d}{a}b & f-\frac{d}{a}c \\ 0 & 0 & i-\frac{g}{a}c -\frac{h-\frac{g}{a}b}{ e-\frac{d}{a}b}(f-\frac{d}{a}c) \end{bmatrix}\sim\begin{bmatrix}a & b & c \\ 0 & e-\frac{d}{a}b & f-\frac{d}{a}c \\ 0 & 0 & \frac{a e i - a f h - b d i + b f g + c d h - c e g}{a e - b d} \end{bmatrix}$

this formula is found:

$ det(B)=a(e-\frac{d}{a}b)(\frac{a e i - a f h - b d i + b f g + c d h - c e g}{a e - b d}) =a e i - a f h - b d i + b f g + c d h - c e g $

I find it is easier to calculate the determinant using these six 'diagonals'.

When this method is used on $(B-λI)$

$C=\begin{bmatrix}a-λ & b & c \\ d & e-λ & f \\ g & h & i-λ \end{bmatrix}\sim \begin{bmatrix}a-λ & b & c \\ 0 & (e-λ)-\frac{d}{a-λ}b & f-\frac{d}{a-λ}c \\ 0 & 0 & (i-λ)-\frac{g}{a-λ}c -\frac{h-\frac{g}{a-λ}b}{ (e-λ)-\frac{d}{a-λ}b}(f-\frac{d}{a-λ}c) \end{bmatrix}$

$Det(C) = \frac{(a - λ) (e - λ) - bd}{(a - λ) (e - λ) - bd}-λ^3 + (a + e + i)λ^2 - (ae + ai + ei )λ+(fh + bd + cg )λ+(aei + bfg + cdh - ceg - bdi- afh)$

meaning that the characterisitic polynomial can be written as:

• $-λ^3$
• $+tr(C)λ^2$
• $-$sum of products of all values on the 'main' diagonal.
• $+$sums of products of all values on the other diagonals.
• $+det(A)$

In my original case:

$-λ^3 + (6+6+5)λ^2 -(36+30+30)λ + (4+1+1)λ + (180-20-6-6-2-2)$

$=-λ^3 + 17λ^2 -90λ + 144$