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The question may sound stupid but I really don't understand the logic behind this. Whenever we do a PCA, we take a covariance matrix on the centered data and do eigen decomposition. In order to project the data to the new eigen basis, in all sources they take a dot product with the eigen vector itself instead of taking an inverse of the eigen vector to project the data to the new eigen basis. *Shouldn't the data be multiplied(dot product) with the inverse of the eigen basis vectors since M = inv(B2)B1, where B2 is the new basis(eigen) and B1 is our basis(standard basis). Since B1 is a standard basis, M becomes M = inv(B2). Why is inverse not taken when projecting the data to new basis? Please explain

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    $\begingroup$ Please explain what the inverse of a vector might be. $\endgroup$
    – whuber
    Commented Nov 22, 2022 at 15:20
  • $\begingroup$ Suppose you have a vector [1 2] and you want to change the vector from standard basis to a different basis, we write the basis vectors in the form of a matrix and take the inverse of the new basis matrix as here B1 will be the standard basis and B2 will be the new basis $\endgroup$ Commented Nov 23, 2022 at 4:29
  • $\begingroup$ The eigenbasis in PCA is orthonormal. That is, for any such basis $(e_1,e_2,\ldots,e_n),$ $e_i\cdot e_j=\delta_{ij}$ (where $\delta_{ij}=1$ iff $i=j$ and otherwise $\delta_{ij}=0.$) When any vector $x$ is expressed in this basis it will be of the form $x = \alpha_1e_1+\alpha_2e_2+\cdots+\alpha_ne_n.$ To find the coefficient $\alpha_j,$ note that $x\cdot e_j=\alpha_1\delta_{1j}+\alpha_2\delta_{2j}+\cdots+\alpha_n\delta_{nj}=\alpha_j.$ $\endgroup$
    – whuber
    Commented Nov 23, 2022 at 14:53

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