# Reversing PCA back to 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:

$PC_1 = X_1L_1 + X_2L_2 + ... + X_nL_n$

My question is, if I pick an arbitrary point in the PCA space (i.e. a value for $PC_1$ and $PC_2$, 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., $X_1,X_2,\dots,X_n$)?

Note 100% reversal is obviously not expected (since I'm only using 2 PCs), so a decent approximation is fine.

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Read the first chapter of Michael Greenacre's Biplots in Practice. Then read the rest of it when you understand that much! It is in essence a demonstration of how to do the type of interpretation from a biplot that you are asking for. –  Andy W Aug 20 '12 at 18:03

Yes. Basically, what you did was to do: $$PC=\mathbf{V}X,$$ where $PC$ are the principal components, $X$ is your matrix with the data 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: $$\mathbf{V}^{-1}PC=X,$$ but, because the matrix of loadings is orthonormal (they have eigenvectors!), then $\mathbf{V}^{-1}=\mathbf{V}^{T}$, so: $$\mathbf{V}^{T}PC=X.$$ 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.

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Ok, but does that apply for any arbitrary point? i.e. a point that wasn't in the original dataset? –  Matt Burland Aug 20 '12 at 19:25
Yes, but it can only be used for interpolation. Suppose datapoint $x_{ab}$ is missing and that the data matrix is centered. The problem is to find $\mathbf{V}$. You can estimate the sample covariance matrix as: $$\hat{\Sigma}_{ij}=\alpha_{ij}\sum_{k}x_{ik} x_{jk},$$ where $$\alpha_{ij}=\left(N-1-\sum_{k}[\delta_{ab}(i,k)+\delta_{ab}(j,k)]\right)^{-1},‌​$$ and $\delta_{ab}(i,k)$ is one when $i=a$ and $k=b$ and zero otherwise. Given this, the transformation matrix $\mathbf{V}$ is straightforward to obtain (you just have to estimate the SVD decomposition of this sample covariance matrix). –  Néstor Aug 20 '12 at 20:22