If D=P−1APD=P^{-1}AP is a diagonalization of T:V→VT:V\to V, why does PP contain the eigenvectors of TT?

My question is: we say that an operator T:V→VT: V\to V is diagonalizable if there exists a corresponding diagonal matrix with respect to some basis of VV. So suppose AA is a matrix corresponding to T with respect to another basis. Then we have some matrix PP such that the diagonal matrix D=P−1APD = P^{-1}AP, where DD contains eigenvalues of TT, and PP is the matrix of eigenvectors. So since P−1APP^{-1}AP is also known as the change of basis matrix, where PP gives the change of basis from one to another, why is that PP contains the eigenvectors of TT?

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“so since P−1APP^{-1}AP is also known as the change of basis matrix” — no. PP is called the change of basis matrix. the matrix P−1APP^{-1}AP is DD, the diagonalized matrix.
– symplectomorphic
Oct 20 at 23:44

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3 Answers
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You have A=PDP−1A=PDP^{-1}. If we let PiP_i denote the iith column of PP and let eie_i denote the iith standard basis vector, then
APi=PDP−1Pi=PDei=Pdiiei=diiPi,AP_i=PDP^{-1} P_i = PDe_i = P d_{ii} e_i = d_{ii} P_i,
which shows that the iith column of PP is a diid_{ii}-eigenvector of TT.

  

 

so I guess my question would be: how would we compute P^(-1)AP if we didnt know P contained the eigenvectors? How would we specifically compute P just from knowing that it is a matrix involved in change of basis?
– asdfasdf
Oct 20 at 23:48

Write PD=APPD = AP, and take eie_i to be the vector whose ii-th coordinate is 11 and all other coordinates are zero, so that PeiPe_i is the ii-th column of PP. Then A(Pei)=PDei=di(Pei)A(Pe_i) = PDe_i = d_i (P e_i), where did_i is the ii-th diagonal entry of DD. This shows that PeiP e_i is an eigenvector with eigenvalue did_i.

This is because if AA is diagonalisable, there exists a basis of eigenvectors. Now PP is precisely the change of basis matrix from the given basis to the basis of eigenvectors, and the matrix of the operator TT in this basis of eigenvectors is given by the change of basis formula:
A′=P−1AP.A’=P^{-1}AP.
Furthermore, by definition of eigenvectors, the matrix in such basis is a diagonal matrix with the corresponding eigenvalues on the diagonal.

  

 

is P not defined the other way around? Where P takes the a vector in terms of basis of eigenvalue and then maps it to vector in terms of other basis (the basis that A is respective to?)
– asdfasdf
Oct 21 at 0:09

  

 

Yes PP expresses the â€کold’ coordinates in function of the â€کnew’ coordinates, and it’s precisely for this reason the â€کnew’ matrix of TT is P−1APP^{-1}AP.
– Bernard
Oct 21 at 0:21

  

 

so can you tell me what the action of P^(-1)AP is on a vector v? So v must be written in terms of eigenvector basis first? then P sends v to v represented by basis of A. A then takes this vector to another vector still represented by the other basis, and finally p^-1 takes it back to a vector written in terms of eigenvectors?
– asdfasdf
Oct 21 at 0:53

  

 

Yes. Just make the abstract calculation, denoting with a ‘ the new coordinates: you have Y=AXY=AX, X=PX′X=PX’ and Y=PY′Y=PY’. So the 1st equality becomes Y=PY′=A(PX′)Y=PY’=A(PX’), whence Y′=P−1A(PX′)=(P−1AP)X′Y’=P^{-1}A(PX’)=(P^{-1}AP)X’.
– Bernard
Oct 21 at 1:05

  

 

So shouldn’t your answer say instead “PP is precisely the change of basis matrix from the eigenvector basis to the (given) basis of AA”?
– Sasho Nikolov
2 days ago