Matrix Eigenvalue formula $|\lambda I-A |$ or $|A-\lambda I|$?

Question

my lecturer wrote a solution for our exam, and he wrote the eigenvalue formula as $|\lambda I-A |$ instead of $|A-\lambda I|$. What I understand is we use the latter to get the eigenvalue of a matrix.

Is this a typo or it doesn't matter whether it's $|A-\lambda I|$ or $|\lambda I-A |$?

Thanks

Answer 1

Both are actually equivalent!

Let us recall how we derive the notion of eigenvalues and such. That is, for a square matrix $A$ and eigenvalues $\lambda$ of $A$, we have:

$$A\vec{x} = \lambda \vec{x}$$

Then,

$$A\vec{x} - \lambda \vec{x} = 0 \;\;\; \Rightarrow \;\;\; (A-\lambda I) \vec{x} = 0$$

Then we try to find $\lambda$ such that $det(A - \lambda I) = 0$.

But wait! We can do this a different way, as:

$$A\vec{x} = \lambda \vec{x} \;\;\; \Rightarrow \;\;\; 0 = \lambda \vec{x} - A \vec{x} \;\;\; \Rightarrow \;\;\; 0 = (\lambda I - A) \vec{x}$$

and thus we seek $\lambda$ such that $det(\lambda I - A) = 0$

Thus, as a result, you can use either equation - $det(\lambda I - A) = 0$ or $det(A - \lambda I) = 0$. Which you use is a matter of personal taste; I was actually taught the former but found the latter more intuitive.

Answer 2

If $A$ is a $n\times n$ matrix, then the polynomials $\det(A-\lambda\operatorname{Id})$ and $\det(\lambda\operatorname{Id}-A)$ are equal if $n$ is even and symmetric otherwise. Therefore, they have the same zeros. So, as far as searching for eigenvalues is concerned, both provide the same information.

Answer 3

Just note that $\det (A-\lambda I) =(-1)^n\det (\lambda I-A)$, so $\lambda$ solves $ \det (A-\lambda I) =0$ if and only if it solves $\det (\lambda I-A)=0$, therefore you can calculate the eigenvalues of $A$ by solving $\det (\lambda I-A)=0$ or $\det (A-\lambda I)=0$. The solutions are the same.