Where does the distinction between $det(\lambda I - A)$ and $det(A - \lambda I)$ for char. poly. of A come from?

Question

I see certain textbooks and people use one or the other and i did the arithmetic on one very basic case and they happenned to come out the same, but I'm still curious where does the difference originate and if it is just the matter of preference?

I first encountered $det(\lambda I - A)$ when reading about the Cayley-Hamilton theorem and was wondering for a bit then whether the inversion has anything to do with the fact that the characteristic polynomial is now(in the context of CH) being calculated with a matrix as an argument instead of a scalar.

Thanks so much.

Answer 1

It is indeed just a matter of preference; whether you're using a scalar or matrix argument makes no difference. Note that if $p(\lambda) = \det(\lambda I - A)$, then $$ \det(A - \lambda I) = \det((-1) \cdot (\lambda I - A)) = (-1)^n \det (\lambda I - A) = (-1)^n p(\lambda) .$$

Answer 2

Some definitions of the characteristic polynomial require it to be monic: the highest-degree term must have a coefficient of $1$. $\det(\lambda I-A)$ guarantees this regardless of the size of $A$. With $\det(A-\lambda I)$, the highest-degree term has a coefficient of $-1$ for odd-order $A$. The two polynomials of course have the same roots since $\det(\lambda I-A)=(-1)^n\det(A-\lambda I)$, where $n$ is the order of $A$.

If the characteristic polynomial doesn’t have to be monic, then this is largely a matter of preference. As a practical matter, $A-\lambda I$ might be slightly better if you’re calculating eigenvalues by hand since you’re less likely to make a silly sign error when forming this matrix.