TITLE: In spectral decomposition/eigenvalue decomposition, why must $T$ be the set of eigenvectors?
QUESTION [0 upvotes]: If there exists a $T$ such that $T^{-1}AT$ is diagonal, then $A$ is
  diagonalizable and $T^{-1}AT = \Lambda = diag(\lambda_1, \cdots, \lambda_n)$

In the claim it says $T$ is any matrix, but in most examples, $T$ is taken as the columns of eigenvectors...
why the loss of generality here?
Does anyone know what operation it is called when you stack all these eigenvectors into a matrix column by column and why eigenvectors in particular?

REPLY [0 votes]: Because that is the definition of an eigenvector : a vector which changes by a prescribed factor (the eigenvalue). To do that, first calculate the length of that vector (${\bf T}^{-1}$), change it by it's eigenvalue $(\text{diag}(\lambda_1,\cdots,\lambda_n))$ and then go back to the original description of the vector ($\bf T$).
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