Why is $\det (A-\lambda I)=0$?

Since you want non-trivial solutions to $(A - \lambda I)x = 0$, you want $(A - \lambda I)$ to be non-invertible (otherwise, its invertible and you get $x = (A - \lambda I)^{-1} \cdot 0 = 0$ which is a trivial solution). But a linear transformation or a matrix is non-invertible if and only if its determinant is $0$. So $\det(A - \lambda I) = 0$ for non-trivial solutions.

Hint: The determinant being zero is equivalent to some other property about the transformation.

Hmm, let's start from here:

We want $Ax = \lambda x$. Hence $(A-\lambda I)x=0$ Now, can you see why we intend the transformation $A-\lambda I$ to have non-trivial solutions? (that's why we set $det(A-\lambda I)$=0)

Remark: If $Tx=0$, when do we have non-trivial solutions? If $T$ is invertible, then $T^{-1} T x = 0$ and $x=0$. So, $T$ shouldn't be invertible. What condition on T can tell us that T is not invertible? How does this relate to $det(A-\lambda I)=0$ in that case?