Diagonalizable vs full rank vs nonsingular (square matrix)

Solution 1:

Full rank and diagonalizability are independent from one another.

1. A diagonalizable matrix with full rank: $$ \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} $$

2. A non diagonalizable matrix with full rank: $$ \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix} $$

3. A diagonalizable matrix without full rank: $$ \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} $$

4. A non diagonalizable matrix without full rank: $$ \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix} $$

Solution 2:

$1.$

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-

1. Characteristic polynomial splits or not,
2. $n-rank(M-\lambda_iI)=$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?