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Subsection 5.2.3 Existence of the LU factorization

Now that we have an algorithm for computing the LU factorization, it is time to talk about when this LU factorization exists (in other words: when we can guarantee that the algorithm completes).

We would like to talk about the existence of the LU factorization for the more general case where \(A \) is an \(m \times n \) matrix, with \(m \geq n \text{.}\) What does this mean?

Definition 5.2.3.1.

Given a matrix \(A \in \C^{m \times n} \) with \(m \geq n \text{,}\) its LU factorization is given by \(A = L U \) where \(L \in \C^{m\times n}\) is unit lower trapezoidal and \(U \in \C^{n \times n} \) is upper triangular with nonzeroes on its diagonal.

The first question we will ask is when the LU factorization exists. For this, we need another definition.

Definition 5.2.3.2. Principal leading submatrix.

For \(k \leq n \text{,}\) the \(k \times k \) principal leading submatrix of a matrix \(A \) is defined to be the square matrix \(A_{TL} \in \C^{k \times k} \) such that \(A = \FlaTwoByTwo{A_{TL}}{A_{TR}}{A_{BL}}{A_{BR}} \text{.}\)

This definition allows us to state necessary and sufficient conditions for when a matrix with \(n \) linearly independent columns has an LU factorization:

Homework 5.2.3.1.

Prove Lemma 5.2.3.3.

Hint

You may use the fact that a triangular matrix has an inverse if and only if it has no zeroes on its diagonal.

Solution
  • (\(\Rightarrow\)): Assume \(A = L U \) is nonsingular. Hence \(A^{-1} exists \text{.}\) Also, since \(L \) is unit lower triangular, it has no zeroes on its diagonal and hence \(L^{-1}\) exists. Now,

    \begin{equation*} U ( A^{-1} L ) = L^{-1} L U A^{-1} L = L^{-1} A A^{-1} L = I. \end{equation*}

    Hence \(U \) has an inverse and is thus nonsingular.

  • (\(\Leftarrow\)): Assume \(A = L U \) and \(U \) has no zeroes on its diagonal. We conclude \(U^{-1} exists \text{.}\) Also, since \(L \) is unit lower triangular, it has no zeroes on its diagonal and hence \(L^{-1}\) exists. Now,

    \begin{equation*} A (U^{-1} L^{-1}) = L U U^{-1} L^{-1} = I. \end{equation*}

    Hence \(A \) has an inverse and is thus nonsingular.

  • (\(\Rightarrow\)): Let nonsingular \(A \) have a (unique) LU factorization. We will show that its principal leading submatrices are nonsingular.

    Let

    \begin{equation*} \begin{array}[t]{c} \underbrace{ \FlaTwoByTwo { A_{TL} }{ A_{TR} } { A_{BL} }{ A_{BR} } } \\ A \end{array} = \begin{array}[t]{c} \underbrace{ \FlaTwoByTwo { L_{TL} }{ 0 } { L_{BL} }{ L_{BR} } } \\ L \end{array} \begin{array}[t]{c} \underbrace{ \FlaTwoByTwo { U_{TL} }{ U_{TR} } { 0 }{ U_{BR} } } \\ U \end{array} \end{equation*}

    be the LU factorization of \(A \text{,}\) where \(A_{TL}, L_{TL}, U_{TL} \in \Ckxk \text{.}\) By the assumption that \(L U \) is the LU factorization of \(A \text{,}\) we know that \(U\) cannot have a zero on the diagonal and hence is nonsingular. Now, since

    \begin{equation*} \begin{array}{rcl} \begin{array}[t]{c} \underbrace{ \FlaTwoByTwo { A_{TL} }{ A_{TR} } { A_{BL} }{ A_{BR} } } \\ A \end{array} \amp = \amp \begin{array}[t]{c} \underbrace{ \FlaTwoByTwo { L_{TL} }{ 0 } { L_{BL} }{ L_{BR} } } \\ L \end{array} \begin{array}[t]{c} \underbrace{ \FlaTwoByTwo { U_{TL} }{ U_{TR} } { 0 }{ U_{BR} } } \\ U \end{array} \\ \amp = \amp \FlaTwoByTwo { L_{TL} U_{TL} }{ L_{TL} U_{TR} } { L_{BL} U_{TL} }{ L_{BL} U_{TR} + L_{BL} U_{BR} }, \end{array} \end{equation*}

    the \(k \times k \) principal leading submatrix \(A_{TL} \) equals \(A_{TL} = L_{TL} U_{TL} \text{,}\) which is nonsingular since \(L_{TL} \) has a unit diagonal and \(U_{TL} \) has no zeroes on the diagonal. Since \(k \) was chosen arbitrarily, this means that all principal leading submatrices are nonsingular.

  • (\(\Leftarrow\)): We will do a proof by induction on \(n \text{.}\)

    • Base Case: \(n = 1 \text{.}\) Then \(A \) has the form \(A = \FlaTwoByOneSingleLine{ \alpha_{11} }{ a_{21} }\) where \(\alpha_{11} \) is a scalar. Since the principal leading submatrices are nonsingular \(\alpha_{11} \neq 0 \text{.}\) Hence \(A = \begin{array}[t]{c} \underbrace{ \FlaTwoByOneSingleLine{ 1 }{ a_{21}/\alpha_{11} } }\\ L \end{array} \begin{array}[t]{c} \underbrace{ \alpha_{11} } \\ U \end{array}\) is the LU factorization of \(A \text{.}\) This LU factorization is unique because the first element of \(L \) must be \(1 \text{.}\)

    • Inductive Step: Assume the result is true for all matrices with \(n = k \text{.}\) Show it is true for matrices with \(n = k+1\text{.}\)

      Let \(A \) of size \(n = k+1 \) have nonsingular principal leading submatrices. Now, if an LU factorization of \(A\) exists, \(A = L U \text{,}\) then it would have to form

      \begin{equation} \begin{array}[t]{c} \underbrace{ \left( \begin{array}{c | c} A_{00} \amp a_{01} \\ \hline a_{10}^T \amp \alpha_{11} \\ A_{20} \amp a_{21} \end{array} \right)} \\ A \end{array} = \begin{array}[t]{c} \underbrace{ \left( \begin{array}{c | c} L_{00} \amp 0 \\ \hline l_{10}^T \amp 1 \\ L_{20} \amp l_{21} \end{array} \right)} \\ L \end{array} \begin{array}[t]{c} \underbrace{ \left( \begin{array}{c | c} U_{00} \amp u_{01} \\ \hline 0 \amp \upsilon_{11} \end{array} \right)} \\ U \end{array} .\label{chapter05-eqn-LUproof2}\tag{5.2.1} \end{equation}

      If we can show that the different parts of \(L\) and \(U \) exist, are unique, and \(\upsilon_{11} \neq 0 \text{,}\) we are done (since then \(U \) is nonsingular). (5.2.1) can be rewritten as

      \begin{equation*} \left( \begin{array}{c } A_{00} \\ \hline a_{10}^T \\ A_{20} \end{array} \right) = \left( \begin{array}{c } L_{00} \\ \hline l_{10}^T \\ L_{20} \end{array} \right) U_{00} \mbox{ and } \left( \begin{array}{c } a_{01} \\ \hline \alpha_{11} \\ a_{21} \end{array} \right) = \left( \begin{array}{c } L_{00} u_{01} \\ \hline l_{10}^T u_{01} + \upsilon_{11} \\ L_{20} u_{01} + l_{21} \upsilon_{11} \end{array} \right), \end{equation*}

      or, equivalently,

      \begin{equation*} \left\{ \begin{array}{rcl} L_{00} u_{01} \amp = \amp a_{01} \\ \nu_{11} \amp = \amp \alpha_{11} - l_{10} u_{01} \\ l_{21} \amp = \amp (a_{21} - L_{20} u_{01} )/\nu_{11} \end{array} \right. \end{equation*}

      Now, by the Induction Hypothesis \(L_{11} \text{,}\) \(l_{10}^T \text{,}\) and \(L_{20} \) exist and are unique. So the question is whether \(u_{01} \text{,}\) \(\upsilon_{11} \text{,}\) and \(l_{21} \) exist and are unique:

      • \(u_{01} \) exists and is unique. Since \(L_{00} \) is nonsingular (it has ones on its diagonal) \(L_{00} u_{01} = a_{01} \) has a solution that is unique.

      • \(\upsilon_{11} \) exists, is unique, and is nonzero. Since \(l_{10}^T \) and \(u_{01} \) exist and are unique, \(\upsilon_{11} = \alpha_{11} - l_{10}^T u_{01} \) exists and is unique. It is also nonzero since the principal leading submatrix of \(A \) given by

        \begin{equation*} \FlaTwoByTwo { A_{00} }{ a_{01} } { a_{10}^T }{ \alpha_{11} } = \FlaTwoByTwo { L_{00} }{ 0 } { l_{10}^T }{ 1} \FlaTwoByTwo { U_{00} }{ u_{01} } { 0 }{ \upsilon_{11} }, \end{equation*}

        is nonsingular by assumption and therefore \(\upsilon_{11} \) must be nonzero.

      • \(l_{21} \) exists and is unique. Since \(\upsilon_{11} \) exists, is unique, and is nonzero,

        \begin{equation*} l_{21} = ( a_{21} - L_{20} a_{01} )/ \upsilon_{11} \end{equation*}

        exists and is uniquely determined.

      Thus the \(m \times (k+1) \) matrix \(A \) has a unique LU factorization.

    • By the Principal of Mathematical Induction the result holds.

The formulas in the inductive step of the proof of Theorem 5.2.3.4 suggest an alternative algorithm for computing the LU factorization of a \(m \times n \) matrix \(A \) with \(m \geq n \text{,}\) given in Figure 5.2.3.5. This algorithm is often referred to as the (unblocked) left-looking algorithm.

\begin{equation*} \newcommand{\routinename}{A = \mbox{LU-left-looking}( A )} \newcommand{\guard}{ n( A_{TL} ) \lt n( A ) } \newcommand{\partitionings}{ A \rightarrow \FlaTwoByTwo{A_{TL}}{A_{TR}} {A_{BL}}{A_{BR}} } \newcommand{\partitionsizes}{ A_{TL} {\rm ~is~} 0 \times 0 } \newcommand{\repartitionings}{ \FlaTwoByTwo{A_{TL}}{A_{TR}} {A_{BL}}{A_{BR}} \rightarrow \FlaThreeByThreeBR{A_{00}}{a_{01}}{A_{02}} {a_{10}^T}{\alpha_{11}}{a_{12}^T} {A_{20}}{a_{21}}{A_{22}} } \newcommand{\moveboundaries}{ \FlaTwoByTwo{A_{TL}}{A_{TR}} {A_{BL}}{A_{BR}} \leftarrow \FlaThreeByThreeTL{A_{00}}{a_{01}}{A_{02}} {a_{10}^T}{\alpha_{11}}{a_{12}^T} {A_{20}}{a_{21}}{A_{22}} } \newcommand{\update}{ \begin{array}{l} \mbox{Solve } L_{00} u_{01} = a_{01} \mbox{ overwriting } a_{01} \mbox{ with } u_{01} \\ \alpha_{11} := \upsilon_{11} = \alpha_{11} - a_{10}^T a_{01} \\ a_{21} := a_{21} - A_{20} a_{01} \\ a_{21} := l_{21} = a_{21} / \alpha_{11} \\ \end{array} } \FlaAlgorithm \end{equation*}
Figure 5.2.3.5. Left-looking LU factorization algorithm. \(L_{00} \) is the unit lower triangular matrix stored in the strictly lower triangular part of \(A_{00} \) (with the diagonal implicitly stored).
Homework 5.2.3.2.

Show that if the left-looking algorithm in Figure 5.2.3.5 is applied to an \(m \times n \) matrix, with \(m \ge n \text{,}\) the cost is approximately \(m n^2 - \frac{1}{3} n^3 \) flops (just like the right-looking algorithm).

Solution

Consider the iteration where \(A_{TL} \) is (initially) \(k \times k \text{.}\) Then

  • Solving \(L_{00} u_{01} = a_{21} \) requires approximately \(k^2 \) flops.

  • Updating \(\alpha_{11} := \alpha_{11} - a_{10}^T a_{01} \) requires approximately \(2k \) flops, which we will ignore.

  • Updating \(a_{21} := a_{21} - A_{20} a_{01} \) requires approximately \(2(m-k-1) k \) flops.

  • Updating \(a_{21} := a_{21} / \alpha_{11} \) requires approximately \((m-k-1) \) flops, which we will ignore.

Thus, the total cost is given by, approximately,

\begin{equation*} \sum_{k=0}^{n-1} \left( k^2 + 2 ( m-k-1 ) k \right) \mbox{ flops}. \end{equation*}

Let us now simplify this:

\begin{equation*} \begin{array}{l} \sum_{k=0}^{n-1} \left( k^2 + 2 ( m-k-1 ) k \right) \\ ~~~=~~~~ \lt \mbox{ algebra } \gt \\ \sum_{k=0}^{n-1} k^2 + 2 \sum_{k=0}^{n-1} ( m-k-1 ) k \\ ~~~=~~~~ \lt \mbox{ algebra } \gt \\ \sum_{k=0}^{n-1} 2 (m-1) k - \sum_{k=0}^{n-1} k^2 \\ ~~~ \approx ~~~~ \lt \sum_{j=0}^{n-1} j \approx n^2/2 \mbox{ and } \sum_{j=0}^{n-1} j^2 \approx n^3/3 \gt \\ (m-1) n^2 + \frac{2}{3} n^3 \\ ~~~=~~~~ \lt \mbox{ simplify } \gt \\ (m-1) n^2 - \frac{1}{3} n^3 \\ ~~~ \approx ~~~ \lt \mbox{ for large } m, (m-1) \approx m \lt m n^2 - \frac{1}{3} n^3. \end{array} \end{equation*}

Had we not ignored the cost of \(\alpha_{11} := \alpha_{11} - a_{10}^T a_{01} \text{,}\) which approximately \(2k \text{,}\) then the result would have been approximately

\begin{equation*} m n^2 - \frac{1}{3} n^3, \end{equation*}

which is identical to that of the right-looking algorithm in Figure 5.2.2.1. Of course, regardless,

\begin{equation*} (m-1) n^2 - \frac{1}{3} n^3. \approx m n^2 - \frac{1}{3} n^3 \end{equation*}

if \(m \) is large.

Remark 5.2.3.6.

A careful analysis would show that the left- and right-looking algorithms perform the exact same operations with the same elements of \(A \text{,}\) except in a different order. Thus, it is no surprise that the costs of these algorithms are the same.

Ponder This 5.2.3.3.

If \(A \) is \(m \times m \) (square!), then yet another algorithm can be derived by partitioning \(A \text{,}\) \(L \text{,}\) and \(U \) so that

\begin{equation*} A = \left( \begin{array}{c c} A_{00} \amp a_{01} \\ a_{10}^T \amp \alpha_{11} \end{array} \right), L = \left( \begin{array}{c c} L_{00} \amp 0 \\ l_{10}^T \amp 1 \end{array} \right), U = \left( \begin{array}{c c} U_{00} \amp u_{01} \\ 0 \amp \upsilon_{11} \end{array} \right). \end{equation*}

Assume that \(L_{00} \) and \(U_{00} \) have already been computed in previous iterations, and determine how to compute \(u_{01} \text{,}\) \(l_{10}^T \text{,}\) and \(\upsilon_{11} \) in the current iteration. Then fill in the algorithm:

\begin{equation*} \newcommand{\routinename}{A = \mbox{LU-left-looking}( A )} \newcommand{\guard}{ n( A_{TL} ) \lt n( A ) } \newcommand{\partitionings}{ A \rightarrow \FlaTwoByTwo{A_{TL}}{A_{TR}} {A_{BL}}{A_{BR}} } \newcommand{\partitionsizes}{ A_{TL} {\rm ~is~} 0 \times 0 } \newcommand{\repartitionings}{ \FlaTwoByTwo{A_{TL}}{A_{TR}} {A_{BL}}{A_{BR}} \rightarrow \FlaThreeByThreeBR{A_{00}}{a_{01}}{A_{02}} {a_{10}^T}{\alpha_{11}}{a_{12}^T} {A_{20}}{a_{21}}{A_{22}} } \newcommand{\moveboundaries}{ \FlaTwoByTwo{A_{TL}}{A_{TR}} {A_{BL}}{A_{BR}} \leftarrow \FlaThreeByThreeTL{A_{00}}{a_{01}}{A_{02}} {a_{10}^T}{\alpha_{11}}{a_{12}^T} {A_{20}}{a_{21}}{A_{22}} } \newcommand{\update}{ \begin{array}{l} ~~ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ ~~ \\ ~~ \\ ~~ \\ \end{array} } \FlaAlgorithm \end{equation*}
This algorithm is often called the bordered LU factorization algorithm.

Next, modify the proof of Theorem 5.2.3.4 to show the existence of the LU factorization when \(A \) is square and has nonsingular leading principal submatrices.

Finally, show that this bordered algorithm also requires approximately \(2 m^3 / 3 \) flops.

Homework 5.2.3.4.

Implement the algorithm given in Figure 5.2.3.5 as

function [ A_out ] = LU_left_looking( A )

by completing the code in Assignments/Week05/matlab/LU_left_looking.m. Input is an \(m \times n \) matrix \(A\text{.}\) Output is the matrix \(A \) that has been overwritten by the LU factorization. You may want to use Assignments/Week05/matlab/test_LU_left_looking.m to check your implementation.

Solution

See Assignments/Week05/answers/LU_left_looking.m. (Assignments/Week05/answers/LU_left_looking.m)