About a matrix A = 2 a 1 0
I want to find a condition for A to be diagonalizable. I think that A needs to have 2 linearly independent eigenvectors, and also Eigenvectors associated to distinct eigenvalues are linearly independent. But I don't know how can I access to this one.
I assume this matrix $A$ would be $\begin{bmatrix} 2 & a \\ 1 & 0 \end{bmatrix}$. You first need to find the eigenvalues ($\lambda$) for this matrix such that $AX = \lambda X$, where $X$ is a vector $\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ and $\lambda$ is a scalar. You can solve for $\lambda$ by manipulating $AX = \lambda X$, where $AX = \lambda X \Rightarrow AX - \lambda X = 0 \Rightarrow (A - \lambda I)X = 0$. $\lambda$ will have nonzero solutions as long as the determinant of $(A-\lambda I)$ is equal to 0, so you can solve for $\lambda$ with the equation det$(A- \lambda I) = 0$. Once you have values of $\lambda$, the eigenvector(s) for each value of $\lambda$ can be found by solving for $X$ in $(A - \lambda I)X = 0$ (remember $X = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$). If you find two distinct, linearly independent eigenvectors, then you can diagonalize $A$ by constructing a matrix $P$ from its eigenvectors, where $D$ (the diagonalized matrix) = $P^{-1}AP$.
$A$ would be non-diagonalizable if (assuming you're working with real numbers) $\lambda$ has a complex result (say, $\lambda = \sqrt{-1}$). It would also be non-diagonalizable if you only found one unique eigenvalue. If the one eigenvalue yielded a singular, constant value of $X$, there is only one possible eigenvector, which is not enough. But even if it yielded infinite distinct eigenvectors, they would need to be multiples of each other-- none of these would be linearly independent from one another, meaning $A$ couldn't have a complete basis of eigenvectors and is therefore, not diagonalizable.