Why is $\lambda I - A$ singular if $\lambda$ is an eigenvalue of $A$? Taken from Wikipedia, this is what's got me confused:

I'm not understanding how v being non-zero implies that its multiplier $(\lambda I - A)$ must be singular. 
Also, I didn't quite get why the roots of the determinant of $(\lambda I - A)$ are the eigenvalues for $A$.
I looked at this post for some hints and I did some searching, but didn't really find a good answer. 
As a side note, I'm going to go buy a linear algebra textbook today as soon as the bookstore opens. I realize this is fundamental stuff that I should already know. 
Thanks in advance.
 A: I'd like to mention that there are many characterisations of the singularity of a matrix (see for instance here). 
If $\lambda I - A$ is singular, this means that there is non-zero vector $v$ such that $(\lambda I - A)v=0$, which can be rewritten as $Av = \lambda v$. Hence $v$ is an eigenvector. 
Also, $\lambda I - A$ being singular is equivalent with $\det(\lambda I - A)=0$. Combining this with the reasoning hereabove shows that roots of $\det(\lambda I - A)=0$ are eigenvalues.
A: $v$ is a null vector of $\lambda I-A$.
You only get that if the RREF has a free variable.
So the RREF has at most $n-1$ pivots, and therefore a row of zeros.  So its determinant is zero.
Each elementary rowop changes the determinant by a nonzero factor so $\lambda I-A$ must have started with zero determinant.
If $(\lambda I-A)v=0$ then expand that, so $Av=\lambda v$
A: Then buy that Linear Algebra textbook, because my answer is based upon basic Linear algebra stuff:


*

*If $A$ is a square matrix, then $A$ is singular if and only if there is a non-zero vector $\mathbf v$ such that $A.\mathbf{v}=0$.

*$\lambda$ is an eigenvalue of $A$ $\iff$ $\lambda\operatorname{Id}-A$ is singular $\iff$ $\det(\lambda\operatorname{Id}-A)=0$.

A: Theorem. An $n \times n$ matrix $A$ is invertible if and only if $Ax = 0$ has only the trivial solution.
Proof. Suppose $A$ is invertible. To solve $Ax = 0$, we pre multiply the equation sequentially by elementary matrices to get the system $Rx = 0$, where $R$ is a row-reduced echelon matrix. Since elementary matrices are invertible, and products of invertible matrices are invertible, we see that $A$ is invertible if and only if $R$ is invertible. But a square row-reduced echelon matrix is invertible if and only if it is the identity, in which case $Rx = 0$ has only the trivial solution $x = 0$.
Conversely, if $Ax = 0$ has only the trivial solution, then so does $Rx = 0$ where $R$ is the row reduced echelon form of $A$, and $Rx = 0$ has the trivial solution if and only if $R$ is the identity matrix, and $R = I$ if and only if $A$ is invertible. $\tag*{$\blacksquare$}$

This is why if $(\lambda I - A)v = 0$ for a nonzero vector $v$, then $\lambda I - A$ must be non-invertible.

Consider the equation $f(x) = \det(xI - A)$. This is a polynomial equation, and if $\lambda$ is an eigenvalue then the above discussion shows that $\det(\lambda I - A) = 0$, or in other words that $\lambda$ is a root of $f$. Conversely, if $\lambda$ is a root of $f$ then $\det(\lambda I - A) = 0$, so $\lambda I - A$ is singular, so $(\lambda I - A)v = 0$ has a solution for a nonzero $v$. In other words, $\lambda$ is an eigenvalue of $A$ with eigenvector $v$.
