Why is it that the characteristic polynomial for a matrix $A$
$$\phi(\lambda) = \det(\lambda I - A)$$
when finding the roots gives the eigenvalues of $A$?
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By definition, an eigenvalue of a matrix $A$ is a number $\lambda$ such that $Ax=\lambda x$ where $x$ is a non zero vector, called the eigenvector corresponding to $\lambda$. This means $(A-\lambda I)x=0$ for this $x$. That is, $\lambda$ is an eigenvalue of $A$ iff $A-\lambda I$ is not invertible. That is, $\lambda$ is an eigenvalue of $A$ iff $\det (A-\lambda I)=0$. Here $\det (A-\lambda I)$ is a polynomial in $\lambda$ and any $\lambda$'s satisfying this polynomial are eigenvalues!
$\lambda$ is an eigenvalue of $A$ iff $A-\lambda I$ has a non-trivial null space, which is true iff $\det(A-\lambda I)=0$, which is equivalent to $p(\lambda)=0$, where $p$ is the characteristic polynomial of $A$.
We need a non zero vector $V$ to satisfy $$AV=\lambda V$$
Thus the homogeneous system $$(A-\lambda I)V=0$$ must have a nontrivial solution.
Thus we have to have $$det(A-\lambda I)=0$$
We want to find all $\lambda $ satisfuing $Ax= \lambda x \rightarrow (\lambda x -Ax)=0 \rightarrow(\lambda I -A)x=0 ; x \neq 0$ Therefore $ \lambda I -A$ is singular, meaning has determinant $0$ . Expanding on the determinant, we find all $\lambda$ that satisfy this condition. Was this clear/helpful?