For example, in R, the MASS::mvrnorm() function is useful for generating data to demonstrate various things in statistics. It takes a mandatory Sigma argument which is a symmetric matrix specifying the covariance matrix of the variables. How would I create a symmetric $n\times n$ matrix with arbitrary entries?


Create an $n\times n$ matrix $A$ with arbitrary values

and then use $\Sigma = A^T A$ as your covariance matrix.

For example

n <- 4  
A <- matrix(runif(n^2)*2-1, ncol=n) 
Sigma <- t(A) %*% A
  • $\begingroup$ Likewise, Sigma <- A + t(A). $\endgroup$ – rsl May 31 '16 at 8:45
  • 5
    $\begingroup$ @MoazzemHossen: Your suggestion will produce a symmetric matrix, but it may not always be positive semidefinite (e.g. your suggestion could produce a matrix with negative eigenvalues) and so it may not be suitable as a covariance matrix $\endgroup$ – Henry May 31 '16 at 10:30
  • $\begingroup$ Yes, I noticed that R returns error in the event my suggested way produced unsuitable matrix. $\endgroup$ – rsl May 31 '16 at 18:32
  • 4
    $\begingroup$ Note that if you prefer a correlation matrix for better interpretability, there is the ?cov2cor function, which can be applied subsequently. $\endgroup$ – gung May 31 '16 at 22:58
  • 1
    $\begingroup$ @B11b: You need your covariance matrix to be positive semi-definite. That would put some limits on the covariance values, not totally obvious ones when $n \gt 2$ $\endgroup$ – Henry Jan 11 '18 at 9:26

I like to have control over the objects I create, even when they might be arbitrary.

Consider, then, that all possible $n\times n$ covariance matrices $\Sigma$ can be expressed in the form

$$\Sigma= P^\prime\ \text{Diagonal}(\sigma_1,\sigma_2,\ldots, \sigma_n)\ P$$

where $P$ is an orthogonal matrix and $\sigma_1 \ge \sigma_2 \ge \cdots \ge \sigma_n \ge 0$.

Geometrically this describes a covariance structure with a range of principal components of sizes $\sigma_i$. These components point in the directions of the rows of $P$. See the figures at Making sense of principal component analysis, eigenvectors & eigenvalues for examples with $n=3$. Setting the $\sigma_i$ will set the magnitudes of the covariances and their relative sizes, thereby determining any desired ellipsoidal shape. The rows of $P$ orient the axes of the shape as you prefer.

One algebraic and computing benefit of this approach is that when $\sigma_n \gt 0$, $\Sigma$ is readily inverted (which is a common operation on covariance matrices):

$$\Sigma^{-1} = P^\prime\ \text{Diagonal}(1/\sigma_1, 1/\sigma_2, \ldots, 1/\sigma_n)\ P.$$

Don't care about the directions, but only about the ranges of sizes of the the $\sigma_i$? That's fine: you can easily generate a random orthogonal matrix. Just wrap $n^2$ iid standard Normal values into a square matrix and then orthogonalize it. It will almost surely work (provided $n$ isn't huge). The QR decomposition will do that, as in this code

n <- 5
p <- qr.Q(qr(matrix(rnorm(n^2), n)))

This works because the $n$-variate multinormal distribution so generated is "elliptical": it is invariant under all rotations and reflections (through the origin). Thus, all orthogonal matrices are generated uniformly, as argued at How to generate uniformly distributed points on the surface of the 3-d unit sphere?.

A quick way to obtain $\Sigma$ from $P$ and the $\sigma_i$, once you have specified or created them, uses crossprod and exploits R's re-use of arrays in arithmetic operations, as in this example with $\sigma=(\sigma_1, \ldots, \sigma_5) = (5,4,3,2,1)$:

Sigma <- crossprod(p, p*(5:1))

As a check, the Singular Value decomposition should return both $\sigma$ and $P^\prime$. You may inspect it with the command


The inverse of Sigma of course is obtained merely by changing the multiplication by $\sigma$ into a division:

Tau <- crossprod(p, p/(5:1))

You may verify this by viewing zapsmall(Sigma %*% Tau), which should be the $n\times n$ identity matrix. A generalized inverse (essential for regression calculations) is obtained by replacing any $\sigma_i \ne 0$ by $1/\sigma_i$, exactly as above, but keeping any zeros among the $\sigma_i$ as they were.

  • $\begingroup$ It might help to demonstrate how to use the rows of $P$ to orient the axes as preferred. $\endgroup$ – gung May 31 '16 at 23:33
  • 1
    $\begingroup$ Might be worth mentioning that the singular values in svd(Sigma) will be reordered -- that confused me for a minute. $\endgroup$ – FrankD Mar 28 '18 at 10:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.