# The space of matrices requiring pivoted LU factorization is almost full-dimensional

The pivoted LU factorization of square matrices is the key routine for the direct solution of linear systems. All square, invertible matrices have a pivoted LU factorization of the form $\mathbf{PA} = \mathbf{LU}$, where $\mathbf{P}$ is a permutation matrix and $\mathbf{L}, \; \mathbf{U}$ are respectively upper and lower triangular.

However, not all square, invertible matrices have a factorization of the form $\mathbf{A} = \mathbf{LU}$ (an LU factorization). For example, consider the invertible matrix

$$\mathbf{A} = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}$$

for which the unpivoted LU algorithm will attempt to divide by 0 on the first iteration. We will call such matrices “pivot-required”, meaning that they do not have a “pure” LU factorization $\mathbf{A} = \mathbf{LU}$ (even in infinite precision).

An interesting question regarding such matrices is: what is the dimension of the set of pivot-required matrices?

The answer, using elementary means is: the set of pivot required matrices has dimension $n^2 - 1$.

The full proof is presented below, but it is instructive to look at the $2 \times 2$ case first. First, we see that the zero must be encountered in the upper left on the first iteration; since we are assuming the matrix is invertible, a zero will not be encountered on the second (read last) iteration. Now, the set of full rank $2 \times 2$ matrices with a specified element equal to zero is 3-dimensional, and the generic part of this set admits the parameterization

$$\begin{bmatrix} 0 & b \\ a & c \end{bmatrix}$$

which only fails to be full rank on sets of dimension 2 or 0 like

$$\begin{bmatrix} 0 & 0 \\ a & c \end{bmatrix}, \; \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} .$$

Since these “voids” in the set of pivot-required matrices are of dimension less than 3, the set of pivot-required $2 \times 2$ matrices has dimension 3.

Note that here $3 = 2^2 - 1$, which agrees with the general formula stated above and proven below.

Proof

As above, one can explicitly parameterize the space of $n \times n$ matrices for which unpivoted LU fails on the first iteration; now one finds that the dimension of this space is $n^2 - 1$.

However, there exist pivot-required matrices for which unpivoted LU fails on the $k$th iteration; for example, the following matrix has a failure on the second iteration:

$$\begin{bmatrix} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix} .$$

In general, pivot-required matrices can cause failure on iterations $1, ..., n-1$, since generating a zero on the last iteration would imply the matrix is not full rank.

Therefore, we must ensure that the set of matrices which fail on iterations $2, ..., n-1$ also have dimension less than or equal to $n^2 - 1$. In fact, we will prove constructively that such sets of matrices are also of dimension exactly $n^2 - 1$.

During the $j$ iteration of the unpivoted LU algorithm the following update is made to the trailing $n-j \times n-j$ submatrix

$$\mathbf{B}_{j+1:,j+1:} \leftarrow \mathbf{B}_{j+1:,j+1:} - \frac{1}{\mathbf{B}_{jj}} \mathbf{B}_{j+1:,j} \mathbf{B}_{j,j+1:} \; .$$

Let $b_{ij}^{(p)}$ denote the $ij$th element of this submatrix after $p$ steps of the unpivoted LU algorithm. Using the update rule for the complete matrix, we can write an update rule corresponding to iteration $p$ for a single element

$$b_{ij}^{(p)} = b_{ij}^{(p-1)} - \frac{1}{b_{pp}^{(p-1)}} b_{il}^{p-1} b_{lj}^{p-1}$$

noting two important facts: $b_{ij}^{(0)} = a_{ij}$ (an element of the original matrix), this update formula only applies to elements in the submatrix $(i,j > p)$.

The condition for breakdown of the algorithm on the $k$th iteration is $b_{kk}^{(k-1)} = 0$. Using the update formula and recurrence relationship, one can equivalently write this condition only in terms of the elements of the original matrix $\mathbf{A}$. Further, applying this recurrence relationship only to the $b_{kk}^{(p)}$ term, one can see that the final result is an equation like $a_{kk} = f(a_{ij})$ where only $i = j = k$ is excluded. Therefore, with only one of the $n^2$ variables dependent on the others, one can explicitly parameterize any such pivot-required matrix with $n^2 - 1$ parameters, hence this space is of dimension $n^2 - 1$.

• The main result presented here implies the following statement: generic $$n \times n$$ matrices have an unpivoted LU factorization in infinite precision.