How to define $\lambda$ as an eigenvalue of an $n \times n$ matrix

Question

I do not exactly understand what this question is asking.

So, it is given to let $A$ be a $n \times n$ matrix. Then, I must "define that $\lambda$ is an eigenvalue of this matrix.

I currently understand that in order to solve for a matrix's eigenvalues (at least how my professor describes it), one must use the following equation: $$ 0 = \det(A- \lambda I) $$ How does one go about proving this? I do not understand how i can take the determinant this matrix, let alone solve the equation above.

Answer 1

Given a square matrix $A$ we say $\lambda$ an eigen value for $A$ if associated with $\lambda$ atleast one vector $v \neq 0$ such that

$$Av = \lambda v$$.

Suppose that the eigen value $\lambda$ is known then we have a nonzero $v$ such that

$$(A-\lambda I)v=0$$.

Since we assumed $v$ is non zero the new matrix $A_{\lambda}:= A-\lambda I$ is not invertible. Thus we must have

$$\det A_\lambda=\det (A-\lambda I)=0$$.