LU Decomposition Explained Why Matrix E Fails And Equations For Matrix A

Why does the matrix E = ((0, 1), (1, 1)) not have an LU decomposition? If the matrix A = (aij)n×n has an LU decomposition, determine the n^2 equations with n^2 unknowns.

In the realm of linear algebra, matrix decomposition is a fundamental technique used to simplify complex matrix operations. One of the most widely used matrix decompositions is the LU decomposition, which factors a matrix into the product of a lower triangular matrix (L) and an upper triangular matrix (U). This decomposition is particularly valuable for solving systems of linear equations, computing determinants, and inverting matrices. However, not all matrices possess an LU decomposition. This article delves into why a specific matrix, denoted as E, lacks an LU decomposition and explores the general equations involved in the LU decomposition of a matrix A.

Why Matrix E Fails to Have LU Decomposition

Let's consider the matrix

E = \begin{pmatrix}
0 & 1 \\
1 & 1
\end{pmatrix}

To understand why matrix E does not have an LU decomposition, we attempt to find matrices L and U such that E = LU, where L is a lower triangular matrix and U is an upper triangular matrix. Suppose we assume the following forms for L and U:

L = \begin{pmatrix}
l_{11} & 0 \\
l_{21} & l_{22}
\end{pmatrix}, U = \begin{pmatrix}
u_{11} & u_{12} \\
0 & u_{22}
\end{pmatrix}

Then, the product LU is:

LU = \begin{pmatrix}
l_{11} & 0 \\
l_{21} & l_{22}
\end{pmatrix} \begin{pmatrix}
u_{11} & u_{12} \\
0 & u_{22}
\end{pmatrix} = \begin{pmatrix}
l_{11}u_{11} & l_{11}u_{12} \\
l_{21}u_{11} & l_{21}u_{12} + l_{22}u_{22}
\end{pmatrix}

For E to equal LU, we must have:

l_{11}u_{11} = 0 \\
l_{11}u_{12} = 1 \\
l_{21}u_{11} = 1 \\
l_{21}u_{12} + l_{22}u_{22} = 1

The first equation, l11u11 = 0, implies that either l11 = 0 or u11 = 0. If l11 = 0, then the second equation, l11u12 = 1, cannot be satisfied because 0 * u12 cannot equal 1. Similarly, if u11 = 0, the third equation, l21u11 = 1, cannot be satisfied because l21 * 0 cannot equal 1. Therefore, no such L and U matrices exist that satisfy E = LU. This demonstrates that matrix E does not have an LU decomposition in its original form.

To obtain an LU decomposition, we can perform a row exchange on matrix E to obtain a new matrix where the element in the first row and first column is non-zero. By swapping the rows of E, we get:

P = \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}

PA = \begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}

Now, PA has an LU decomposition. This simple example illustrates that the existence of an LU decomposition can depend on the arrangement of the matrix elements, specifically the presence of a zero in the pivot position (the first element in the first row).

General Equations in LU Decomposition of Matrix A

Now, let's consider a general n × n matrix A and explore the equations that arise when we attempt to decompose it into L and U matrices. Suppose A = (aij)n×n, L = (lij)n×n is a lower triangular matrix, and U = (uij)n×n is an upper triangular matrix. For the LU decomposition to exist, we need A = LU. This means:

(a_{ij})_{n \times n} = (l_{ij})_{n \times n} (u_{ij})_{n \times n}

When we multiply L and U, we obtain a set of n^2 equations. Each equation corresponds to an element in the resulting matrix, which must be equal to the corresponding element in A. The matrix multiplication gives us:

a_{ij} = \sum_{k=1}^{n} l_{ik}u_{kj}

However, since L is a lower triangular matrix and U is an upper triangular matrix, we have the conditions:

l_{ik} = 0 \text{ for } k > i

u_{kj} = 0 \text{ for } k > j

Therefore, the sum can be simplified based on these conditions. Specifically:

a_{ij} = \sum_{k=1}^{\min(i, j)} l_{ik}u_{kj}

This gives us n^2 equations in n^2 unknowns (the elements of L and U). To see this more clearly, let's break it down into cases:

Case 1: i ≤ j (Upper Triangle and Diagonal)

For elements in the upper triangle and diagonal of A (i ≤ j), the equation becomes:

a_{ij} = \sum_{k=1}^{i} l_{ik}u_{kj} = l_{i1}u_{1j} + l_{i2}u_{2j} + \cdots + l_{ii}u_{ij}

Case 2: i > j (Lower Triangle)

For elements in the lower triangle of A (i > j), the equation becomes:

a_{ij} = \sum_{k=1}^{j} l_{ik}u_{kj} = l_{i1}u_{1j} + l_{i2}u_{2j} + \cdots + l_{ij}u_{jj}

Number of Equations

There are n^2 elements in matrix A, which means we have n^2 equations. These equations relate the elements of A to the elements of L and U. The number of unknowns is also n^2 because L and U each have n rows and n columns, but due to the triangular nature of L and U, many of their elements are zero, and the diagonal elements are often normalized (e.g., the diagonal elements of L are set to 1).

Determining the Unknowns

To determine the n^2 unknowns, we have n(n+1)/2 elements in L (including the diagonal) and n(n+1)/2 elements in U (including the diagonal). However, the equations are not necessarily independent, and we need additional constraints to solve for the unknowns uniquely. One common approach is to set the diagonal elements of L (lii) to 1, which is known as Doolittle's decomposition. Alternatively, we can set the diagonal elements of U (uii) to 1, which is known as Crout's decomposition.

Example: 3x3 Matrix

For a 3x3 matrix A, the equations are:

\begin{pmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{pmatrix} = \begin{pmatrix}
l_{11} & 0 & 0 \\
l_{21} & l_{22} & 0 \\
l_{31} & l_{32} & l_{33}
\end{pmatrix} \begin{pmatrix}
u_{11} & u_{12} & u_{13} \\
0 & u_{22} & u_{23} \\
0 & 0 & u_{33}
\end{pmatrix}

Multiplying L and U gives us the following system of equations:

a_{11} = l_{11}u_{11} \\
a_{12} = l_{11}u_{12} \\
a_{13} = l_{11}u_{13} \\
a_{21} = l_{21}u_{11} \\
a_{22} = l_{21}u_{12} + l_{22}u_{22} \\
a_{23} = l_{21}u_{13} + l_{22}u_{23} \\
a_{31} = l_{31}u_{11} \\
a_{32} = l_{31}u_{12} + l_{32}u_{22} \\
a_{33} = l_{31}u_{13} + l_{32}u_{23} + l_{33}u_{33}

This system consists of 9 equations in 9 unknowns. To solve this, we can set the diagonal elements of L to 1 (Doolittle's method) or the diagonal elements of U to 1 (Crout's method) to provide additional constraints.

Conclusion

In summary, the matrix E presented in this article does not have an LU decomposition due to the presence of a zero in the pivot position, which leads to a contradiction when attempting to solve for the elements of L and U. However, row exchanges can be performed to obtain a matrix that does have an LU decomposition. For a general n × n matrix A, the LU decomposition involves solving n^2 equations in n^2 unknowns, derived from the matrix equation A = LU. These equations can be simplified using the triangular properties of L and U. Additional constraints, such as normalizing the diagonal elements of L or U, are required to obtain a unique solution. LU decomposition remains a powerful tool in linear algebra, facilitating the solution of linear systems and other matrix operations, provided that the matrix can be appropriately decomposed.

Understanding the conditions under which LU decomposition exists and the equations involved is crucial for applying this technique effectively in various mathematical and computational contexts. The exploration in this article provides a foundational understanding of these concepts, which are essential for anyone working with matrices and linear systems.