# First component of non-centered data

Let $$X$$ be a random vector of dimension $$p$$ and $$\{ X_1, \dots, X_n \}$$ the $$n$$ observations of such vector. Let $$\mathbb{X}$$ be the matrix with rows $$X_k$$ and $$\mathbb{X}_c$$ is the matrix with rows $$X_k - \mu_X$$, where $$\mu_X$$ is the empirical mean of $$X$$, $$\mu_X = \sum_{k=1}^n X_k$$.

In class, the question was if the first eigenvector of $$\mathbb{X}^t \mathbb{X}$$ and $$\mathbb{X}_c^t \mathbb{X}_c$$ are different. The answer is yes and we saw an example like this: Where the blue segment is the first eigenvector of the centered points $$\mathbb{X}_c^t \mathbb{X}_c$$ and the black segment is the first eigenvector of the points without centering $$\mathbb{X}^t \mathbb{X}$$, this vector seems to point to the center of $$X$$.

I know that the matrix $$\mathbb{X}_c^t \mathbb{X}_c$$ is proportional to the covariance matrix $$Cov(X)=\frac{1}{n}\mathbb{X}_c^t \mathbb{X}_c$$, then they have the same eigenvectors. The first eigenvector of $$Cov(X)$$ is the direction of the first principal component.

With some algebra I obtain the following relation between $$Cov(X)$$ and $$\mathbb{X}^t \mathbb{X}$$: $$Cov(X) = \frac{1}{n} \mathbb{X}^t \mathbb{X} - \mu_X \mu_X^t$$

The first eigenvalue of $$\mathbb{X}^t \mathbb{X}$$ is the vector $$v$$ that maximizes $$v^t \mathbb{X}^t \mathbb{X} v$$, then $$v$$ also maximizes: $$v^t ( Cov(X) + \mu_X \mu_X^t ) v$$

My question is if there is an expression of the first eigenvector of $$\mathbb{X}^t \mathbb{X}$$ that could tell me how this vector is going to look geometrically.