Diagonalizing a matrix question on determinant property

Question

So the equation I am used to is $\det(A - \lambda I )x = 0$ when trying to find the eigenspace, but I have seen some people solve for the eigenspace doing $\det(\lambda I - A )$ and was wondering is there a theorem stating that the $\det(A - B) = \det(B - A)$.

Answer 1

In general,

$$\det(-E)=(-1)^n \det(E)$$

where $n$ is the size of the matrix $E$. So $\det(A-B)=(-1)^n\det(B-A)$. Therefore:

$$ \det(\lambda I-A)=0 \iff \det(A-\lambda I)=0 $$ which is what you wanted.