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I am reading a text on principal components which has the following excerpt:

Since $\frac1n \cdot \sum_{i=1}^n x_{ij} =0$, the sample mean of the first principal component scores, $\bar{z}_1$, equals $0$.

The book also mentions:

$$Z_1 =\phi_{11}X_1+\phi_{21}X_2+\ldots+\phi_{p1}X_p$$

I don't understand how the standardization of the predictor variables mandates that the mean of the first principal component scores also be 0. What am I missing? I think part of the challenge is that they didn't provide the exact mathematical formulation of $\bar{z}_1$. What exactly is $\bar{z}_1$?

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  • $\begingroup$ Suppose every data (row) vector $x$ satisfies a set of linear equations $xA=0$ for a fixed matrix $A.$ Since PCs are linear combinations $\omega x$ (for various $\omega,$ one per PC), notice that $(\omega x)A = \omega (xA)=\omega(0)=0,$ too. In other words: the set of solutions of a set of linear equations is a vector subspace. $\endgroup$
    – whuber
    Commented Dec 28, 2021 at 14:20

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PCA creates new variables $Z$ based on linear combinations of the old variables $X$. It is assumed that $X_1, \cdots, X_p$ have been centered (and scaled). In other words, for $j=1,\cdots,p$, $\sum_{i=1}^n x_{ij} = 0$. Let the first principal component equal \begin{eqnarray*} z_{1i} = \sum_{j=1}^p \phi_{j1} x_{ij} \end{eqnarray*} The textbook means that $\bar{z}_1$ equals \begin{eqnarray*} \bar{z}_1 &=& \frac{1}{n}\sum_{i=1}^n z_{1i} \\ &=& \frac{1}{n}\sum_{i=1}^n \sum_{j=1}^p \phi_{j1} x_{ij} \\ &=& \frac{1}{n} \sum_{j=1}^p \phi_{j1} \sum_{i=1}^n x_{ij} \\ &=& 0. \end{eqnarray*}

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  • $\begingroup$ I have used PCA before bur never thought of this point before. I will come back to this answer in a few months when I need to use it again. Great answer! $\endgroup$
    – stats_noob
    Commented Aug 13, 2022 at 15:10
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If the $X_i$'s are zero mean a linear combination of them will also have zero mean. Am I missunderstanging your question?

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