On Diagonalization of a Matrix

Question

When we try to think what a transform looks like in a coordinate system with some other basis using change of basis, we do the following, let the unknown system be A and we know about B

1. We take the vector defined with A's basis, and convert it to our language, using a matrix M (which contains A's basis in B's language)

2. Apply the transform, in B's system

3. Revert back to A's coordinates by multiplying inverse of M

So my question begins here. we go through all the trouble of diagonolization, because we know how the transform will look like, if the basis are Eigen-vectors. So, we convert the vector specified in standard coordinate system to Eigen basis and apply the transform in the Eigen basis and take the inverse transform. I get all this, but what I don't get is that, why is the product of the three matrices $M^{-1}GM=D$ contains the diagonal elements as the Eigen values. Doesn't this product $M^{-1}GM$ give the transformed vector in the standard coordinate system.
This leads me to the question,

1. In what coordinate system is this diagonalized matrix D defined? if its defined in the standard system, it doesn't transform the vector under consideration to the required vector, since the transform considered, $G$ ,is not scaling(which is what the diagonal matrix might do)

2. And if $M^{-1}GM$ is going to have eigen values as diagonal elements, why do all these stuff instead of writing a matrix with eigen values in the diagonal. Inshort

Why does the product $M^{-1}GM$ give D

P.S. I get the math part, but I lack insight. My question is primarily based on 1

Answer 1

Assume that we have $n$ distinct eigenvalues for the $n\times n$ matrix $A$ then we can formulate the eigenvalue equations for all the eigenvalue and eigenvector pairs as

$$Av_i=\lambda_iv_i \text{ for } i=1,2,...,n.$$

We can assemble all these equations into one single matrix equation

$$A[v_1,...,v_n]=[v_1,...,v_n]\text{diag}[\lambda_1,...,\lambda_n]$$

If we introduce $V=[v_1,...,v_n]$ and $\Lambda = \mathrm{diag}[\lambda_1,...,\lambda_n]$ we can rewrite this as

$$AV=V\Lambda$$

or by inverting $V$

$$\Lambda = V^{-1}AV.$$

Hence, $V^{-1}AV$ is diagonal.

Answer 2

The Eigenvectors are such that

$$Am_i=m_i d_i $$ where $d_i$ is the associated Eigenvalue.

Write this relation for all Eigenvectors and pack in a matrix, to get

$$AM=MD.$$

$D$ is a digonal matrix, the effect of which is to multiply every column vector by the corresponding diagonal coefficient.

If the Eigenvectors are linearly independent, the matrix $M$ is invertible and you obtain the Eigendecomposition

$$A=AMM^{-1}=MDM^{-1}.$$

From this equation you also draw

$$D=M^{-1}MD=M^{-1}AM.$$