Reversing PCA back to the original variables I have a set of data that has $n$ samples described by $m$ variables. I do a PCA to reduce it to just 2 dimensions so I can make a nice 2D plot of the data. I understand that the $x,y$ coordinates (i.e., the PCA scores) for the plot are calculated by basically summing the products of the original data (after centering) by the loadings for each variable, so:
$$\mathrm{PC}_1 = X_1L_1 + X_2L_2 + ... + X_mL_m.$$
My question is, if I pick an arbitrary point in the PCA space (i.e. a value for $\mathrm{PC}_1$ and $\mathrm{PC}_2$, or $x$ and $y$ in my plot), is there a convenient way to translate that back to a set of the original values (i.e., $X_1,X_2,\dots,X_m$)?  
Note 100% reversal is obviously not expected (since I'm only using 2 PCs), so a decent approximation is fine.
 A: Yes. Basically, what you did was to do:
$$\mathrm{PC}=\mathbf{V}X,$$
where $\mathrm{PC}$ are the principal components, $X$ is your matrix with the data (centered, and with data points in columns) and $\mathbf{V}$ is the matrix with the loadings (the matrix with the eigenvectors of the sample covariance matrix of $X$). Therefore, you can do:
$$\mathbf{V}^{-1}\cdot\mathrm{PC}=X,$$
but, because the matrix of loadings is orthonormal (they are eigenvectors!), then $\mathbf{V}^{-1}=\mathbf{V}^{T}$, so:
$$\mathbf{V}^T\cdot\mathrm{PC}=X.$$
Note that this gives you exactly the same equation you cite for the recovery of the PCs, but now for the data, and you can retain as many PCS as you like.
A: I have a doubt about the above answers. Since after dimension reduction, we only know 2 principal components, and the rest principal components are abandoned. The projection matrix V is not a square matrix (not completely orthonormal, it is a semi-orthogonal matrix). Suppose n is the number of samples and m is the number of variables. $X$ is a $m$-by-$n$ matrix, $V$ is a 2-by-m matrix (whose rows are the top 2 eigenvectors of the covariance matrix of $X$), PC is a 2-by-n matrix. Then we have PC = VX. Then $VV^T$ is an identity matrix, but $V^TV$ is not. Thus $V^TPC=V^TVX$ cannot give us the exact original matrix $X$, since $V^TV$ is not an identity matrix.
