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Population and Sample PCA · Tutorial 497 of 975

Eigenvectors, Loadings, and Score Coefficients

Learn which quantities define a principal-component score, how to calculate its covariances with the original variables, and when those covariances are reported as loadings.

Intermediate 10 min read

What You'll Learn

  • Identify a covariance-matrix eigenvector as the coefficient vector for a principal-component score.
  • Derive the variance of a score and its covariance with each original variable.
  • Distinguish score coefficients from covariance and correlation loadings.
  • Calculate scores, component variances, and variable-component covariances for a specified PCA model.
  • Apply the same covariance identities to centered sample data.

Three Quantities That Are Easy to Confuse

A principal component can be described in several numerical ways. Its eigenvector gives the coefficients used to form a score; the score is the resulting random variable or observed data column; and the covariances between that score and the original variables describe how the variables relate to it. These quantities are connected, but they are not interchangeable. In particular, an eigenvector entry is not generally the covariance between that variable and its component.

We work with a random vector \(X=(X_1,\ldots,X_p)^T\) whose mean vector \(\mu=E[X]\) and covariance matrix \(\Sigma=\operatorname{Cov}(X)\) are finite. The principal components considered here are those of the covariance matrix. Since \(\Sigma\) is real and symmetric, it has orthonormal eigenvectors. Let \(q_j\) be a unit eigenvector with eigenvalue \(\lambda_j\), so \(\Sigma q_j=\lambda_j q_j\).

Definition: The score of the \(j\)th principal component is the centered linear combination \(Z_j=q_j^T(X-\mu)\). The entries of \(q_j\) are its score coefficients. The vector \(\operatorname{Cov}(X,Z_j)\) contains the covariances of the original variables with that score. When an author calls a quantity a “loading,” check the stated convention: it may mean these covariances, the corresponding correlations, or a rescaled eigenvector.

Centering matters because PCA describes variation around the mean. For an observed vector \(x\), the component score is the number \(q_j^T(x-\mu)\), not the coefficient vector \(q_j\). A score is one value for one observation; the coefficient vector specifies how to calculate that value.

What the Eigenvector Determines

The eigenvector is chosen to describe a direction of variation in the original variable coordinates. Its eigenvalue is the variance of the score in that direction. The following identities also give the covariances with each original variable.

Theorem: Suppose \(E[X]=\mu\), \(\operatorname{Cov}(X)=\Sigma\) is finite, and \(q_j\) is a unit eigenvector of \(\Sigma\) with eigenvalue \(\lambda_j\). For \(Z_j=q_j^T(X-\mu)\), $$ \operatorname{Var}(Z_j)=\lambda_j, \qquad \operatorname{Cov}(X,Z_j)=\lambda_j q_j. $$ If \(q_j,q_k\) are orthogonal eigenvectors, then \(\operatorname{Cov}(Z_j,Z_k)=0\).

Proof. Write \(X_c=X-\mu\), so \(E[X_c]=0\) and \(\operatorname{Cov}(X_c)=\Sigma\). By bilinearity of covariance,

$$ \operatorname{Var}(Z_j) =\operatorname{Var}(q_j^T X_c) =q_j^T\Sigma q_j =q_j^T(\lambda_j q_j) =\lambda_j q_j^Tq_j =\lambda_j, $$

because \(q_j\) is a unit vector. For the vector of covariances, taking covariance with each coordinate of \(X_c\) gives

$$ \operatorname{Cov}(X,Z_j) =E[X_c(q_j^T X_c)] =E[X_cX_c^T]q_j =\Sigma q_j =\lambda_j q_j. $$

Finally,

$$ \operatorname{Cov}(Z_j,Z_k) =q_j^T\Sigma q_k =q_j^T(\lambda_k q_k) =\lambda_k q_j^Tq_k =0, $$

when \(q_j^Tq_k=0\). This proves all three identities. \(\square\)

The second identity is the key distinction: the covariance between variable \(X_i\) and score \(Z_j\) is the \(i\)th entry of \(\lambda_jq_j\), or \(\lambda_j q_{ij}\). It equals the eigenvector entry multiplied by the eigenvalue. The vector \(q_j\) itself consists of score coefficients, not variable-score covariances.

Worked Example: Coefficients and Covariances in a Two-Variable PCA

Suppose the centered variable vector has covariance matrix

$$ \Sigma= \begin{pmatrix} 97/25&96/25\\ 96/25&153/25 \end{pmatrix}, \qquad q_1= \begin{pmatrix}3/5\\4/5\end{pmatrix}, \qquad q_2= \begin{pmatrix}-4/5\\3/5\end{pmatrix}. $$

First check the stated eigenpair for \(q_1\). The vectors are unit and orthogonal, since \(3^2+4^2=25\) and \((3)(-4)+(4)(3)=0\). Direct multiplication gives

$$ \Sigma q_1 = \begin{pmatrix} (97\cdot3+96\cdot4)/125\\ (96\cdot3+153\cdot4)/125 \end{pmatrix} = \begin{pmatrix}675/125\\900/125\end{pmatrix} = \begin{pmatrix}27/5\\36/5\end{pmatrix} =9q_1. $$

Thus \(q_1\) is an eigenvector with eigenvalue \(9\). The corresponding score and its coefficient vector are

$$ Z_1=\frac35(X_1-\mu_1)+\frac45(X_2-\mu_2), \qquad q_1=\begin{pmatrix}3/5\\4/5\end{pmatrix}. $$

The variance of \(Z_1\) is \(9\), while the variable-score covariance vector is

$$ \operatorname{Cov}(X,Z_1)=9q_1 =\begin{pmatrix}27/5\\36/5\end{pmatrix}. $$

So the first variable has score coefficient \(3/5\), but its covariance with the score is \(27/5\). The second variable's coefficient is \(4/5\), while its covariance with the score is \(36/5\). The two pairs differ because the eigenvalue is \(9\).

Covariance Loadings and Correlation Loadings

The phrase “loading” is not used identically in every PCA text or software package. One precise option is to call \(\operatorname{Cov}(X_i,Z_j)=\lambda_j q_{ij}\) the covariance loading of variable \(i\) on component \(j\). Another commonly reported quantity is the correlation between the variable and the score. If \(\operatorname{Var}(X_i)=\sigma_{ii}>0\) and \(\lambda_j>0\), that correlation is

$$ \operatorname{Corr}(X_i,Z_j) =\frac{\operatorname{Cov}(X_i,Z_j)} {\sqrt{\operatorname{Var}(X_i)\operatorname{Var}(Z_j)}} =\frac{\sqrt{\lambda_j}\,q_{ij}}{\sqrt{\sigma_{ii}}}. $$

The formula follows by substituting \(\operatorname{Cov}(X_i,Z_j)=\lambda_jq_{ij}\) and \(\operatorname{Var}(Z_j)=\lambda_j\) into the definition of correlation. Some sources call these correlations “loadings”; others use “loading” for a rescaled coefficient such as \(\sqrt{\lambda_j}q_{ij}\). State the formula being used rather than inferring a meaning from the label alone.

Worked Example: Converting Covariances to Correlations

Continue with the two-variable model above. The marginal variances are \(\operatorname{Var}(X_1)=97/25\) and \(\operatorname{Var}(X_2)=153/25\). The first component has variance \(9\), so its standard deviation is \(3\). Its variable-score covariances are \(27/5\) and \(36/5\).

For the first variable,

$$ \operatorname{Corr}(X_1,Z_1) =\frac{27/5}{\sqrt{97/25}\,3} =\frac{27/5}{( \sqrt{97}/5)3} =\frac{9}{\sqrt{97}}. $$

For the second variable,

$$ \operatorname{Corr}(X_2,Z_1) =\frac{36/5}{\sqrt{153/25}\,3} =\frac{36/5}{(\sqrt{153}/5)3} =\frac{12}{\sqrt{153}} =\frac{4}{\sqrt{17}}. $$

These are correlations, not coefficients: approximately \(0.914\) and \(0.970\), respectively. They differ from \(3/5\) and \(4/5\) because the original variables have different variances. The formula also shows why correlation loadings are often used when interpreting components: they put variable-score association on a common, dimensionless scale.

For PCA of standardized variables, the same identities apply to the standardized vector and its correlation matrix. In that setting each original standardized variable has variance one, so the correlation between it and the score is \(\sqrt{\lambda_j}q_{ij}\). This is not generally equal to \(q_{ij}\). In covariance PCA with variables in different units, the marginal standard deviations must remain in the denominator when calculating correlations.

From a Model to an Observed Score

A score coefficient is a rule applied to centered data; it is not itself an observation. For an observed vector \(x\), subtract the model mean first and take the inner product with the eigenvector. The following example carries out those steps and checks the associated covariance vector.

Worked Example: Calculating a Score for One Observation

Let \(X=(X_1,X_2,X_3)^T\) have mean \(\mu=(1,1,0)^T\) and covariance matrix

$$ \Sigma= \begin{pmatrix} 2&1&0\\ 1&2&0\\ 0&0&1 \end{pmatrix}. $$

For \(q_1=(1,1,0)^T/\sqrt{2}\), multiplication gives

$$ \Sigma q_1 =\frac{1}{\sqrt{2}} \begin{pmatrix}3\\3\\0\end{pmatrix} =3q_1. $$

Thus the first score is \(Z_1=((X_1-1)+(X_2-1))/\sqrt{2}\), with variance \(3\). For the observed vector \(x=(4,0,2)^T\), first center:

$$ x-\mu=(3,-1,2)^T. $$

Then the observed score is

$$ q_1^T(x-\mu) =\frac{3+(-1)+0\cdot2}{\sqrt{2}} =\sqrt{2}. $$

The coefficient vector remains \((1,1,0)^T/\sqrt{2}\); \(\sqrt{2}\) is this observation's score. Separately, the variable-score covariance vector is

$$ \operatorname{Cov}(X,Z_1)=\Sigma q_1=3q_1 =\begin{pmatrix}3/\sqrt{2}\\3/\sqrt{2}\\0\end{pmatrix}. $$

It describes population covariances, not the score of the particular observation.

The Same Identities for Sample PCA

For sample PCA, let \(x_1,\ldots,x_n\) be \(p\)-dimensional observations, let \(\bar{x}\) be their sample mean, and define the centered data matrix \(X_c\) to have rows \((x_i-\bar{x})^T\). With sample covariance \(S=X_c^TX_c/(n-1)\), suppose \(Sq_j=\hat{\lambda}_j q_j\), where \(q_j\) is a unit eigenvector. The vector of observed scores is \(t_j=X_cq_j\). Directly,

$$ \frac{t_j^Tt_j}{n-1} =q_j^TSq_j =\hat{\lambda}_j, \qquad \frac{X_c^Tt_j}{n-1} =Sq_j =\hat{\lambda}_jq_j. $$

The first expression is the sample variance of the score column; the second gives the sample covariances of the original variables with that score. These are the sample counterparts of the population identities. If a different covariance divisor is used, the eigenvalues and covariance calculations must be interpreted with that same convention.

Common Mistake and Why the Distinction Matters

Wrong reasoning: “The first eigenvector is \((3/5,4/5)^T\), so the covariances of the two variables with the first score are \(3/5\) and \(4/5\).” This treats score coefficients as variable-score covariances. Correction: multiply the unit eigenvector by its eigenvalue: with eigenvalue \(9\), those covariances are \(27/5\) and \(36/5\). If correlations are wanted instead, divide each covariance by the product of the corresponding standard deviations.

Keeping these quantities separate prevents errors in reporting and interpretation. Coefficients tell how a score is constructed; the score gives a coordinate for an observation; covariance loadings quantify variable-score covariation in the variables' original units; and correlation loadings standardize that association. Always identify which quantity is being reported and use its defining formula.

Check Your Understanding

Use the definitions and identities in this tutorial to answer the following.

  1. If \(q_j\) is a unit eigenvector with eigenvalue \(\lambda_j\), what is the variance of \(q_j^T(X-\mu)\)?
  2. Why is \(q_{ij}\) generally not equal to \(\operatorname{Cov}(X_i,Z_j)\)?
  3. For an observed vector \(x\), write the formula for its \(j\)th component score and explain why it is not the eigenvector.
  4. What additional quantities are needed to convert a variable-score covariance into a correlation?
  5. In sample PCA, what do \(t_j=X_cq_j\) and \(X_c^Tt_j/(n-1)\) represent?
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