Why is eigendecomposition $V \Lambda V^{-1}$ not $V^{-1} \Lambda V$ Say you have a linear transformation matrix $A$. In the basis of eigenvectors, this transformation simply becomes a scaling, represented by the diagonal matrix of eigenvalues.
Thus, intuitively the transformation A can be decomposed into the following:


*

*Transform into the basis of eigenvectors (using the transformation matrix $V$, where the eigenvectors form the columns)

*Apply the scaling.

*Transform back.


This would seem to correspond to $V^{-1} \Lambda  V$, where the standard notation of matrices being applied on the left and vectors on the right holds.
Yet everywhere I always see the formula as $V \Lambda  V^{-1}$. Why isn't my intuition correct?
 A: It all depends on what you define your coordinate transformation matrix $V$ to be; obviously if you replace it by the inverse matrix (which carries the same information) then the two possible formulae for the diagonalisation are interchanged. Now typically people take $V$ to be matrix whose columns contain the coordinates of a chosen basis of eigenvectors, the coordinates being expressed of course in terms of the basis for which the matrix $A$ was originally expressed. And it is a sad fact of life that multiplying by that matrix will perform the coordinate transformation in the opposite sense, in other words convert a vector expressed in coordinates on the basis of eigenvectors to its expression in the original basis. Think of it: if you apply $V$ to a standard basis vector, the result is a column of $V$, and therefore will express an eigenvector (one whose coordinates with respect to the eigenvector basis are given by that standard basis vector) in coordinates with respect to the original basis.
A: The formula was generated from the equation $AV=V\Lambda$ which is a compact way of presentation a set of formulas $Av_i=\lambda_i{v_i}$ for eigenvectors.
In this case matrix $\Lambda$ is scaling column vectors ${v_i}$ grouped in the matrix $V=[v_1 \ \  v_2 \ \dots \ \ v_n]$.
