Diagonalizable vs full rank vs nonsingular (square matrix)
Full rank and diagonalizability are independent from one another.
A diagonalizable matrix with full rank: $$ \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} $$
A non diagonalizable matrix with full rank: $$ \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix} $$
A diagonalizable matrix without full rank: $$ \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} $$
A non diagonalizable matrix without full rank: $$ \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix} $$
$1.$
No. A full rank matrix implies it's determinant is non-zero or the matrix is non-singular.
Speaking in terms of linear operator $T$ over a vector space $V$ is diagonalizable if and only if there exists an ordered basis consisting of eigenvectors of $T$, Furthermore if $T$ is diagonalizable and $\beta$ is an ordered basis of $T$ consisting eigen vectors of $T$, then $M=[T]_\beta$, the matrix representation of $T$, is a diagonal matrix.
One can check if a given matrix $M_{n\times n}$ is diagonalizable or not by-
- Characteristic polynomial splits or not,
- $n-rank(M-\lambda_iI)=$multiplicity of $\lambda_i$.
If this two criteria fulfills then a matrix is diagonalizable.
Counterexample: $\begin{bmatrix}3 & 1 &0 \\ 0&3&0\\ 0&0&4\end{bmatrix}$
$2$. Consider the identity matrix $I$, all it's rows and columns are linearly independent. What can you say about the eigenvectors?