# geometric interpretation of deviation vector in direction of an eigenvector (of a covariance matrix)

Let $$D = \begin{bmatrix} x_{11} & \cdot \cdot \cdot & x_{1d}\\ \cdot \cdot \cdot & \\ \cdot \cdot \cdot & \\ \cdot \cdot \cdot &\\ x_{N1} & \cdot \cdot \cdot & x_{Nd} \end{bmatrix} = \begin{bmatrix} X_{1}\\ \cdot \cdot \cdot \\ \cdot \cdot \cdot \\ \cdot \cdot \cdot \\ X_{N} \end{bmatrix}$$

whose columns (or variables) are $$\begin{bmatrix} x_{11}\\ \cdot \cdot \cdot \\ \cdot \cdot \cdot \\ \cdot \cdot \cdot \\ x_{N1} \end{bmatrix}, \cdot \cdot \cdot ,\begin{bmatrix} x_{1d}\\ \cdot \cdot \cdot \\ \cdot \cdot \cdot \\ \cdot \cdot \cdot \\ x_{Nd} \end{bmatrix}$$

Let the mean associated with each variable in the data set D be given by the row vector $$\bar{\mu} = \begin{bmatrix} \bar{\mu_{1}} &\cdot \cdot \cdot & \bar{\mu_{d}} \end{bmatrix}$$

Suppose that the covariance matrix associated with the dataset $$D$$ yields a set $$E$$ of eigenvectors (column vectors) $$E = {E_{1}, ..., E_{d}}$$, all of which are mutually orthogonal since the symmetry of the covariance matrix guarantees the orthogonality of the set $$E$$ and whose corresponding eigenvectors are $$\lambda_{1}, \cdot \cdot \cdot \lambda_{d}$$

What does, for a data point $$X$$,

$$\sum_{j = 1}^{d}\frac{|X - \bar{\mu}|\cdot\vec{E_{j}}}{\lambda_{j}}$$

tells me? particularly, I am interested to know what the projection of $$|X - \bar{\mu}|$$ in the direction of an arbitrary $$j^{th}$$ eigenvector tells me.

• A more standard linear combination of the eigenvectors can be found in this closely related answer. Dec 8, 2021 at 11:21