determinantmatrixsingular
Singular matrix is defined as a square matrix with determinant of zero. I am aware that linear dependency among columns or rows leads to determinant being equal to zero (e.g. one column is a linear composite of other columns).
I am interested whether other conditions exist, except for linear dependency, that can lead to determinant of 0 in Singular Matrix.
Excuse me if I am missing on something obvious, as I am somewhat of a beginner.
My guess is that the numbers are too big (the determinant is large) and you're running into a computational problem.
I was able to replicate your error by running:
> X <- cbind(1,exp(rexp(100,rate=1/50)))
> det(t(X) %*% X)
[1] 5.156683e+126
> solve(t(X) %*% X)
> Error in solve.default...
The problem is numerical. You might be able to solve it by making some transformation of your $X$ matrix that makes the numbers smaller but allows you to work out what $\left(X'X\right)^{-1}$ is.
By setting $\tau=0.99$ you are essentially fitting to the noise at the extreme high-end of your sampling distribution. If you imagine the only-slightly-more-extreme case with $\tau=1.0$, you would be fitting exclusively to the points comprising the upper-envelope of your data, in other words sparsely-distributed points which are almost guaranteed to be unrepresentative outliers. See the plot below for reference.
What is it about your specific use-case that makes you want to perform your fit at such a high percentile?
Best Answer
I think you are operating on an incorrect understanding about the relationship between linear dependence and singularity. Many matrices are not square, and thus do not have a determinant, yet they can have columns that are linearly dependent or independent. In general, if the columns of the matrix $\mathbf{x}$ are linearly dependent then the determinant of the Gramian matrix of $\mathbf{x}$ is zero. That is, you have:
$$\det (\mathbf{x}^\text{T} \mathbf{x}) = 0 \quad \quad \iff \quad \quad \text{columns of matrix } \mathbf{x} \text{ are linearly dependent}.$$
This relationship holds for matrices of any dimension. However, in the special case where $\mathbf{x}$ is a square matrix, you then have $\det (\mathbf{x}^\text{T} \mathbf{x}) = (\det \mathbf{x})^2$, which means that:
$$\quad \quad \quad \quad \det (\mathbf{x}) = 0 \quad \quad \iff \quad \quad \text{columns of square matrix } \mathbf{x} \text{ are linearly dependent}.$$