# In spectral decomposition/eigenvalue decomposition, why must $T$ be the set of eigenvectors?

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?

• Multiply the equation by $T$, you end up with $AT = \Lambda T \Rightarrow A T_j = \lambda_j T_j$ for all $j$. The 'loss of generality' is (I think) up in the orthogonality of $T$. – stochasticboy321 Sep 16 '15 at 2:20

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 ($\bf T$).