Sea \(Z\) la componente principal obtenida de la tranformación lineal de una matriz \(X\), de \(nxp\), y un vector \(a\).

\[Z = \begin{pmatrix} x_{11} & \cdots & x_{1p} \\ \vdots & \ddots & \vdots \\ x_{n1} & \cdots & x_{np} \end{pmatrix} \left[\begin{array}{@{}c@{}} a_{11} \\ \vdots \\ a_{p1} \end{array} \right] = \left[\begin{array}{@{}c@{}} \lambda_{1} \\ \vdots \\ \lambda_{n} \end{array} \right]\\\]

La varianza de la componente quedaría:

\[VAR(Z) = \frac{1}{n}\sum_{i=1}^{n}(\lambda_{i})^{2} = \frac{1}{n} [\lambda_{1}, ..., \lambda_{n}] \left[\begin{array}{@{}c@{}} \lambda_{1} \\ \vdots \\ \lambda_{n} \end{array} \right] = \left. \frac{1}{n}Z^{t}Z \right\rvert_{Z = Xa} = \left. \frac{1}{n}(Xa)^{t}Xa \right\rvert_{(Xa)^{t} = a^{t}X^{t}} = \frac{1}{n}a^{t}X^{t}Xa = a^{t}(\frac{1}{n}X^{t}X)a = a^{t}\Sigma a\\\]

Si no se exige expresamente que el módulo de \(a\) sea 1, entonces podemos maximizar la varianza de la componente principal multiplicando por un escalar, \(\alpha\):

\[a' = \alpha a\\\] \[Z' = Xa' = X\alpha a = \left[\begin{array}{@{}c@{}} \alpha\lambda_{1} \\ \vdots \\ \alpha\lambda_{n} \end{array} \right]\\\] \[VAR(Z') = \frac{1}{n}\sum_{i=1}^{n}(\alpha\lambda_{i})^{2} = \frac{1}{n}(\alpha)^{2}\sum_{i=1}^{n}(\lambda_{i})^{2} = \left. (\alpha)^{2}\frac{1}{n}\sum_{i=1}^{n}(\lambda_{i})^{2} \right\rvert_{\frac{1}{n}\sum_{i=1}^{n}(\lambda_{i})^{2} = VAR(Z)}= \alpha^{2}VAR(Z)\\\]