I am following a book where the "diagonalizability" has been introduced as follows:
Consider a basis formed by a linearly independent set of eigen vectors
$\{v_1,v_2,\dots,v_n\}$. Then it is claimed that with respect to this
basis, the matrix $A$ is diagonal.
I am confused at the word "basis" here. In some other books it is said that a matrix $A$ is called diagonalizable if there exists matrix $P$ such that $P^{-1}AP$ is a diagonal matrix.
I don't think that diagonalizability has anything to do with basis. It just happened that the set of eigen vectors form a basis of $\Bbb R^n$. Also I don't think having a set of $n$ linearly independent eigen vectors is a necessary condition for a matrix to be diagonalizable.
Best Answer
A complex matrix $A\in\mathbb{C}^{n\times n}$ is diagonalizable iff its eigenvectors form a basis of $\mathbb{C}^n$ (note that even real matrices can have complex eigenvectors). That is there exists a nonsingular matrix $P$ such that $P^{-1}AP=D$, where $D$ is diagonal containing the eigenvalues of $A$ with corresponding eigenvectors being the columns of $P$. Also, $P^{-1}AP=D$ means that $AP=PD$, that is, $A$ acts in the basis formed by columns $P$ as diagonal matrix ($D$).
Some notes:
Not all matrices are diagonalizable. For example nontrivial Jordan blocks: $$ A = \begin{bmatrix} 1 & 1 & & & & \\ & 1 & 1 & & & \\ & & \ddots & \ddots & & \\ & & & 1 & 1 \\ & & & & 1 \end{bmatrix}\in\mathbb{R}^{n\times n} $$ has 1 eigenvalue of multiplicity $n$ but with only one (linearly independent) eigenvector.
A matrix with distinct eigenvalues is diagonalizable.