Is there a geometrical interpretation to the notion of eigenvector and eigenvalues?

Question

The wiki article on eigenvectors offers the following geometrical interpretation:

Each application of the matrix to an arbitrary vector yields a result which will have rotated towards the eigenvector with the largest eigenvalue.

Qn 1: If there is any other geometrical interpretation particularly in the context of a covariance matrix?

The wiki also discusses the difference between left and right eigenvectors.

Qn 2: Do the above geometrical interpretations hold irrespective of whether they are left or right eigenvectors?

Answer 1

Here is a partial answer in the case where M is a real symmetric matrix. This is to ensure, by the real spectral theorem, that M has real eigenvectors with real eigenvalues, so there is a chance for a genuine geometric interpretation which stays in $R^n$.

M acts on the unit sphere in $R^n$ in the following way: it sends the unit sphere $v^T v = 1$ to $v^T (M^T M) v = 1$ . This modified shape is not generally a sphere, but is generally an ellipsoid. The axes of this ellipsoid are the eigenvectors of M, and the sizes of each axis are given by the squares of the corresponding eigenvalues.

Answer 2

Of course! Consider a coordinate transformation of rotation and/or scaling (but not translation):

v = Au

where v and u are vectors, and A is a transformation matrix. Then the eigenvectors, if they have real components, are the axes which are left unrotated (scaling only) by the transformation. (see wikipedia)

A covariance matrix is a symmetric, positive definite matrix, so it has orthonormal eigenvectors, and these form a tuple of axes; I am fairly sure the eigenvectors form a new basis of linear combinations of the input variables where the basis variables are uncorrelated, but I can't remember how to show this.

For example, if w1 = [x;y] is a pair of independent unit-variance zero-mean Gaussian random variables, consider w2 = [u;v] = [1 1; 2 1][x;y] = (x+y,2x+y), so that w1 = [-1 1;2 -1][u;v] = [v-u;2u-v]. Then cov(w2) = [2 3; 3 5]. This has eigenvectors which have sqrt(5) in them, hmmmm...

As for question 2, I'm not sure.

Answer 3

If you are interested in covariance matrices, then the eigenvectors of the covariance matrix tell you how to change variables to make your random variables uncorrelated.

Specifically, let $M$ be a covariance matrix of the random variables $X_1,...,X_n$. For simplicity, lets assume that all of these random variables are zero mean. Lets also define $X$ to be the random vector whose $i$"th component is the random variable $X_i$. Let the eigenvectors of $M$ be $v_1, v_2, ..., v_n$. We can assume that these are orthogonal since $M$ is symmetric. Consider the random variables $Y_i$ obtained by taking the dot product of $v_i$ and $X$.

Then the random variables $Y_1, ..., Y_n$ are uncorrelated!

Indeed:

$$ E[ Y_i Y_j] = E[ v_i^T X X^T v_j] = v_i^T M v_j = \lambda_j v_i^T v_j = 0 $$

Answer 4

Instead of giving an answer, let me point out to you this chapter in Cleve Moler's book "Numerical Computing with MATLAB", there is a nice geometric demonstration in MATLAB on how eigenvalues/eigenvectors (as well as singular values/vectors) of an order-2 square matrix are involved in how a circle is transformed into an ellipse after a linear transformation represented by the matrix.