Why can the determinant be assumed to be 0? I'm trying to work through how to calculate eigenvalues and eigenvectors.

I start with 

$$Ax=\lambda x$$

Where $A$ is a $p \times p$ matrix, $\lambda$ is the eigenvalue and $x$ is the eigenvector.

This is the same as:

$$Ax=I\lambda x$$

$$Ax-I\lambda x=0$$

$$(A-I\lambda) x=0$$

We define the matrix $A$ as a $2 \times 2$ matrix:

$\begin{bmatrix}4 &amp; -2\\-3 &amp; 6\end{bmatrix}$

Thus this -$I\lambda$ equals

$\begin{bmatrix}4-\lambda &amp; -2\\-3 &amp; 6-\lambda\end{bmatrix}$

$$Det(A-I\lambda)=(4-\lambda(6-\lambda)-(-3)*-2)$$

$$Det(A-I\lambda)=24-10\lambda +\lambda^2 -6$$
$$Det(A-I\lambda)=18 - 10\lambda + \lambda^2 $$

Then, out of the blue my textbook claims that

$$0=30 - 10\lambda + \lambda^2 $$

How do I justify setting the determinant to $0$?

(I do "not" have an advanced knowledge in linear algebraic analysis, I only know how the determinant is used to calculate the inverse matrix)
The text is not claiming that the determinant is $0$. The text is saying "Let's find out for which values of lambda the determinant is $0$!"

So the determinant is $\lambda^2 - 10\lambda + 30$, and you want to find the $\lambda$ such that it is equal to zero. What do you do? You set it equal to zero and solve for $\lambda$. That is, you solve the equation

$$\lambda^2 - 10\lambda + 30 = 0$$



As for why you are interested in the values of $\lambda$ that make the determinant equal to $0$, remember that 

$$rank(A-\lambda I) = n \iff det(A - \lambda I) \neq 0$$

So, if $det(A-\lambda I) \neq 0$, you will find that the only solution to $(A - \lambda I)x = 0$ is $x = 0$ (due to the fact that the rank of the matrix is full, hence the kernel only contains the $0$ vector). This means that the only $x$ such that $Ax = \lambda x$ is $x=0$, which means that $x$ is not an eigenvector. 

So the only way to have eigenvectors is to have the determinant of $A - \lambda I$ be equal to zero, so that's why to find eigenvalues you look for the values of $\lambda$ that make $det(A - \lambda I) = 0$
As per the Eigen vectors for a transformation matrix , the transformation on such vector will result in scaling it to some values λ. Let X be the Eigen vector and A be the transformation matrix of X.
Thus, 
AX = λX (λ being the scaling factor)
AX-λX=0
(A-λ*I)*X = 0

Now this (A-λI) can be considered as another transformation matrix on X which needs to be 0.
According to a algebraic rule if a transformation matrix transforms the vector X to zero , then its determinant must be zero. Which is exactly the condition we need to satisfy by finding λ(We call it Eigen values from here on) using det(A-λI)=0.
Thus we make the condition det(A-λ*I) =0 and find λ (Eigen value) from it.
Then supplying back Eigen values we get its corresponding Eigen vectors.

This is something I understood from why we need to use determinant .