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:

  1. Transform into the basis of eigenvectors (using the transformation matrix $V$, where the eigenvectors form the columns)
  2. Apply the scaling.
  3. 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?

  • 1
    $\begingroup$ You intuition would be correct if you calculate the left eigenvector. People tend to like the right eigenvectors more, so that is the "convention". By the way, the equation as it stands is incomplete as you should explain what $\Lambda$ and $V$ are in terms of the eigenvalues and eigenvectors. $\endgroup$ – Fabian May 9 '17 at 4:51
  • $\begingroup$ $Y=V^{-1}$. Now how does it change. what looks more pleasing to you? $\endgroup$ – dineshdileep May 9 '17 at 4:51
  • $\begingroup$ @Fabian What is meant by right vs. left eigenvalues? In the equation $A \vec{v} = \lambda \vec{v}$ would $\lambda$ be a right or left eigenvalue? $\endgroup$ – 1110101001 May 9 '17 at 4:53
  • $\begingroup$ That looks like a right eigenvector to me. The left eigenvector is $\vec{w} A = \lambda \vec{w}$ $\endgroup$ – Fabian May 9 '17 at 4:55
  • $\begingroup$ @Fabian Hm alright after reading up a bit it seems a left eigenvector would then be $\vec{v} A = \lambda \vec{v}$. But I'm still missing some key intuition here. The right eigenvector reads naturally as applying the linear transformation A to $\vec{v}$ and finding that the result is scaled by $\lambda$. Yet for some reason this "natural" interpretation seems to go awry when following the line of reasoning in the opening post. How would you even interpret $\vec{v} A$ — applying a vector on a matrix?? $\endgroup$ – 1110101001 May 9 '17 at 5:00

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.


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 $V=[v_1 \ \ v_2 \ \dots \ \ v_n]$.


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