Matrix Equation $A\vec{v}=\lambda \vec{v} \rightarrow \det(A-\lambda I) = 0$ I'm going over eigenvectors and eigenvalues, and I'm confused about the following:
$A\vec{v}=\lambda \vec{v}$
$A\vec{v}-\lambda I\vec{v} = 0$
$(A-\lambda I)\vec{v}=0$
So far so good, but this step I don't understand:
$\det(A-\lambda I) = 0$
How did $\vec{v}$ just fall off?
Would appreciate some clarification, thanks in advance!
 A: $(A - \lambda I)v = 0$ implies that $v$ is in the null space of $A - \lambda I$. Assuming $v$ is nonzero (which is true if $v$ is an eigenvector), this implies that $A - \lambda I$ is not injective, hence not invertible, hence its determinant is zero.
A: There are various equivalent statements about invertibility of a matrix $B$. Some are

*

*$B$ is invertible

*$\det(B) \ne 0$

*the only solution to $Bv=0$ is $v=0$.

In your case, with $B:=A-\lambda I$ you have $Bv=0$ (where $v$ is assumed to be nonzero, from the definition of an eigenvector), so $B$ cannot be invertible, and thus the determinant must be zero.
A: You lost something: in the first equation, you're supposing that there's a nonzero vector $v$ with the property that $Av = v$. This lets you conclude that $(A-\lambda I)$ sends the nonzero vector $v$ to $0$. Since it also sends $0$ to $0$, transformation defined by multiplication $A - \lambda I$ must be non-injective, hence not invertible.
But if the determinant were NONzero, then the matrix would have an inverse (by Cramer's rule, for instance). Hence the determinant must be zero.
This argument does not work unless you use that $v$ is nonzero, though. You could check that using $A = I$ and $\lambda = 0$, with $v = 0$.
