Why $\det(A-λI)$ can be written as a polynomial about $λ$?

Question

I know this is one of the definitions of eigenvalue. But how can I prove this with only the definition: $$Ax = λx$$?

I am a beginner of LA, so this question may seem stupid. Thank you guys.

Answer 1

It is because it is the composition of two polynomial functions: $\lambda\mapsto A-\lambda\operatorname{Id}$ and $\det$.

Answer 2

If $A$ is an $n\times n$ matrix,$$\det(A-\lambda I)=\sum_\sigma\prod_{i=1}^n(A-\lambda I)_{i\sigma(i)},$$where $\sum_\sigma$ sums over all permutations $\sigma$ of $\{1,\,\cdots,\,n\}$. In terms of the Kronecker delta, $(A-\lambda I)_{ij}=A_{ij}-\delta_{ij}\lambda$. Each of the above products is then a polynomial in $\lambda$, as is the sum.

Answer 3

This follows from the definition of the determinant of a square matrix $(a_{ij}),$ which is always a type of homogeneous polynomial of its entries $a_{ij}.$

For the matrix $A-\lambda I,$ where $A$ is a given square matrix and $I$ is the identity matrix of same order, we see that all the entries are numbers, except for those involving the unknown $\lambda.$ It is therefore not surprising that this is a polynomial in $\lambda.$