So a standard vector $\vec{v} = [\begin{smallmatrix}x \\ y \end{smallmatrix}]$ is really just a linear combination of the basis vectors such that $\vec{v} = x \boldsymbol{ \hat{i}} + y \boldsymbol{\hat{j}}$ where $\boldsymbol{\hat{i}} = [\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}]$ and $\boldsymbol{\hat{j}} = [\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}]$. So what a matrix $A = [\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}]$ means is, where would be vector end up in our traditional basis if I decided to replace $\boldsymbol{\hat{i}}$ and $\boldsymbol{\hat{j}}$ with $[\begin{smallmatrix} a \\ c \end{smallmatrix}]$ and $[\begin{smallmatrix} b \\ d \end{smallmatrix}]$ respectively, meaning $A\vec{v} = x[\begin{smallmatrix} a \\ c \end{smallmatrix}] +y[\begin{smallmatrix} b \\ d \end{smallmatrix}]$.
In the case where $[\begin{smallmatrix} a \\ c \end{smallmatrix}]$ and $[\begin{smallmatrix} b \\ d \end{smallmatrix}]$ are linearly dependent, that means that $A$ projects all of 2-space onto a single line. So here's a question, what would that transformation do to the area of a 2D figure by reducing it to a line? Right, it would reduce the area to $0$. Therefore, if $\det(A) =0$, then $A$ is not diagonizable.
The other instance is when there is only one eigenvector span, e.g. $[\begin{smallmatrix} 1 & 1 \\0 & 1 \end{smallmatrix}]$. That one gets into the eigenvector definition, which actually covers both cases:
$$A\vec{v} = \lambda I \vec{v}$$
$$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \vec{v} = \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} \vec{v}$$
$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \vec{v} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} \vec{v} = \vec0$$
$$\bigg(\begin{bmatrix} a & b \\ c & d \end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix}\bigg) \vec{v} = \vec0$$
$$ \begin{vmatrix} a - \lambda & b \\ c & d-\lambda \end{vmatrix} = 0$$
$$ (a - \lambda)(d-\lambda ) - bc = 0$$
So what you now have is a quadratic equation to determine your eigenvalues. Using factoring (or the quadratic formula if you're super desperate), you can find if you have no real solutions (no eigenvalues - bad), one solution, or rather two identical solutions (one eigenvalue - bad), or two real solutions (two eigenvalues - possible).
If you have two real solutions, you'll want to check your eigenvalues by using them to solve a linear system of equations, to wit:
$$ (a - \lambda) x + by = x$$
$$ cx + (d - \lambda) y = y$$
If those solutions are nontrivial, (i.e. $x \ne 0$ or $y \ne 0$), then congratulations, you have a diagonizable linear transformation matrix in 2-space.
tl;dr
$$ (\det(A) \ne 0) \land (N(\lambda)=2) \land (\lambda = p + 0i | p \in \mathbb{R}) \land (\vec{v}_{\lambda} \ne \vec0) \implies A \text{ is diagonizable}$$