Is there any relationship between 'invertible' and 'diagonalizable'? I'am a linear algebra beginner and I am confusing with invertible and diagonalizable.
From my understanding, invertible means non-singular and any of eigenvalue must not be 0. Diagonalizable means there must be N linearly independent eigenvectors.(Eventhough eigenvalue has 0, it seems possible to have N linearly independent eigenvectors. Right?)
Above are just single fraction of their property, and I cannot imagine bigger picture than this.
Is there any intuitive relation or theorem between 'invertible' and 'diagonalizable'?
 A: 
From my understanding, invertible means non-singular and any of eigenvalue must not be 0. 

Exactly. In fact, a matrix is singular if and only if $0$ is its eigenvalue.

Diagonalizable means there must be N linearly independent eigenvectors.(Eventhough eigenvalue has 0, it seems possible to have N linearly independent eigenvectors. Right?)

Correct. Even if an eigenvalue is $0$, a matrix can have $N$ linearly independent eigenvectors. For example, the zero matrix has $N$ linearly independent eigenvectors, because every vector is an eigenvector for the zero matrix.


Is there any intuitive relation or theorem between 'invertible' and 'diagonalizable'?

Not directly, in the sense that one would imply another. You can have matrices in all four classes, i.e.


*

*Invertible and diagonalizable. An example of this is the idenity matrix $\begin{bmatrix}1&0\\0&1\end{bmatrix}$.

*Invertible and not diagonalizable. The simples example would be $\begin{bmatrix}1&1\\0&1\end{bmatrix}$.

*Singular and diagonalizable. The zero matrix $\begin{bmatrix}0&0\\0&0\end{bmatrix}$. is the simplest example of this.

*Singular and non-diagonalizable. The simplest example is probably $\begin{bmatrix}0&1\\0&0\end{bmatrix}$.

