Why is $\det (A-\lambda I)=0$? I'm not sure I understand the logic behind why $\det (A-\lambda I)=0$ for any non-trivial solution to $(A-\lambda I)x=0$.
 A: 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.
A: Hint: The determinant being zero is equivalent to some other property about the transformation.
A: 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?
A: If there exists a non-trivial $x$ s.t. $(A - \lambda I)x=0$, that implies that the mapping $(A-\lambda I)$ must project the vector $x$ onto $0$.
This could only be the case if $x$ lies in the null space of $(A-\lambda I)$. If $(A-\lambda I)$ has a null space, that implies that its rows are not linearly independent.
We know for matrices with non linearly independent rows that $det(A-\lambda I) = 0$.
(Consider that with linearly dependent rows, we could use matrix row operations to change an entire row of the matrix to zeroes.)
