Diagonliazing matrix Suppose I have a linear operator $T : V \to V : v \mapsto A  v$
I want to find the diagonalized version of $A$.
Why do people don't just calculate the eigenvalues of $A$ and put them on a diagonal?
People always calculate the eigenvectors first, put them in a matrix $P$ and calculate $P^{-1}  A  P$ which seems useless to me because it will always end up with the eigenvalues at the diagonal.
 A: The question is fair. Fix $T: V \to V$, linear, with $\dim V = n$. Take a basis $B = \{b_i\}_{i=1}^n$. Suppose that $T$ is diagonalizable, and fix $E = \{e_i\}_{i=1}^n$ a basis of eigenvectors. 
We have that $[T]_{E}$ is a diagonal matrix. But what's the point of computing this matrix? We want to know easily the value of $Tx$, but we usually have the expression for $x$ in the basis $B$, not in the basis $E$. Explicitly: $$[T]_{B} = [{\rm Id}]_{E,B}[T]_{E}[{\rm Id}]_{B,E}.$$
If you don't go through the hassle of computing $P = [{\rm Id}]_{B,E}$, as above, and do: $$[T]_E [x]_B,$$ you won't get anything meaningful. However, we have: $$[Tx]_E = [T]_E [x]_E \quad\text{and}\quad [Tx]_B = [T]_B[x]_B.$$ The point is: $$[Tx]_B =[T]_B[x]_B =[{\rm Id}]_{E,B}[T]_{E}[{\rm Id}]_{B,E}[x]_B =[{\rm Id}]_{E,B}\color{red}{[T]_{E}[x]_E} =[{\rm Id}]_{E,B}\color{red}{[Tx]_{E}} , $$ where the part in red is easy to compute.
A: Not every matrix is diagonalizable. Furthermore, not every $n\times n$ matrix has $n$ eigenvectors. However, every matrix does have $n$ generalized eigenvectors. So you can always compute a Jordan Normal Form instead. This is why we don't just blindly try to diagonalizable every matrix. 
