LU factorization of a nonsingular matrix exists if and only if all leading principal submatrices are nonsingular.

Question

I'm struggling to prove this theorem. I can prove that if the $LU$ factorization exists, then the leading principal submatrices are nonsingular.

To do that, I can show that the determinant of every leading principal submatrix is not zero. (The leading principal submatrix is the product of $L$ and $U$ corresponding leading principal submatrices , and determinant of every $L$ leading principal submatrix is $1$ and determinant of the $U$ leading principal submatrix is product of the diagonal elements).

To prove that if the leading principal submatrices are nonsingular, then $LU$ factorization exists, I believe I should use induction, but I'm getting nowhere. Can anyone help me with the proof?

Answer 1

We show by induction that every $n \times n$ matrix $A$ with nonsingular leading principal minors has a factorization $A = LU$ where $L$ is strictly lower triangular, $U$ is upper triangular, and $L$ and $U$ are both nonsingular. (This statement, as you show, is an if-and-only-if.)

The $1\times 1$ base case is just factoring $a = 1 \cdot a$. To induct, write your $n \times n$ matrix $A$ as a leading principal $(n-1) \times (n-1)$ matrix $A'$ and some leftover entries: $$ A = \left[\begin{array}{ccc|c} & & & \\ & A' & & \vec{b} \\ & & & \\ \hline & \vec{c}^{\mathsf T} & & d \\\end{array}\right]. $$ By the inductive hypothesis (since all leading principal minors of $A'$ are also leading principal minors of $A$), $A'$ has an $LU$ factorization as $A' = L' U'$ with nonsingular $L'$, $U'$. We want to use this to make the factorization $$ \left[\begin{array}{ccc|c} & & & \\ & A' & & \vec{b} \\ & & & \\ \hline & \vec{c}^{\mathsf T} & & d \\\end{array}\right] = \left[\begin{array}{ccc|c} & & & \\ & L' & & \vec{0} \\ & & & \\ \hline & \vec{x}^{\mathsf T} & & 1 \\\end{array}\right]\left[\begin{array}{ccc|c} & & & \\ & U' & & \vec{y} \\ & & & \\ \hline & \vec{0}^{\mathsf T} & & z \\\end{array}\right] $$ work, by picking appropriate $\vec x$, $\vec y$, and $z$.

By doing the block multiplication, we get four equations.

• We have $A' = L'U' + \vec{0}\vec{0}^{\mathsf T}$, which we know is true, so that's done.
• We have $\vec b = L'\vec y + \vec 0 z$, so we want to set $\vec y = L'^{-1}\vec b$. Fortunately that's possible since $L'$ is invertible.
• We have $\vec c^{\mathsf T} = \vec{x}^{\mathsf T}U' + \vec{0}^{\mathsf T}$, so we want to set $\vec{x}^{\mathsf T} = \vec{c}^{\mathsf T}U'^{-1}$. This is possible since $U'$ is also invertible.
• We have $d = \vec{x}^{\mathsf T}\vec y + z$, so we want to set $z = d - \vec{x}^{\mathsf T} \vec y$.

For future inductive steps, we also want to know that the resulting matrices $L$ and $U$ are nonsingular. This is immediate for $L$ since its diagonal is $1$; for $U$, it's not obvious how to check that the value of $z$ we get is nonzero. But once we have $A = LU$ where $A$ and $L$ are nonsingular, we know that $U = L^{-1}A$ is nonsingular.

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There are also $LU$ factorizations out there for which $U$ is singular (some of the diagonal entries of $U$ are zero). For these, there is not an if-and-only-if condition this nice.

You can see from the above proof, for instance, that if $A$ is possibly singular but all of its proper leading principal minors are still nonsingular, then we get a factorization $A = LU$ in which the bottom right entry is possibly $0$. (This is because arguing $z\ne 0$ is the only place where we needed $A$ to be nonsingular.)

Answer 2

Here is an explicit proof, for the enjoyment of a commenting troll. The following goes back to Gauss, from what I've been told.

Explicit LU-decomposition

Theorems

Notation. Fix a commutative ring $\mathbb{K}$. Let $A = \left(a_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq m} \in \mathbb{K}^{n \times m}$ be an $n\times m$-matrix over $\mathbb{K}$. Let $u_1, u_2, \ldots, u_p$ be any elements of $\left\{1,2,\ldots,n\right\}$. Let $v_1, v_2, \ldots, v_q$ be any elements of $\left\{1,2,\ldots,m\right\}$. Then, $\operatorname{sub}^{v_1, v_2, \ldots, v_q}_{u_1, u_2, \ldots, u_p} A$ shall denote the $p\times q$-matrix $\left(a_{u_i, v_j}\right)_{1\leq i\leq p,\ 1\leq j\leq q} \in \mathbb{K}^{p\times q}$.

(Thus, when $u_1 < u_2 < \cdots < u_p$ and $v_1 < v_2 < \cdots < v_q$, this matrix $\operatorname{sub}^{v_1, v_2, \ldots, v_q}_{u_1, u_2, \ldots, u_p} A$ is the matrix obtained from $A$ by crossing out all rows except for the rows numbered $u_1, u_2, \ldots, u_p$ and crossing out all columns except for the columns numbered $v_1, v_2, \ldots, v_q$. It is called a submatrix of $A$.)

Let $A$ be an $n\times n$-matrix. Define an $n \times n$-matrix $R_A =\left( b_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq n}$ by \begin{equation} b_{i,j}=\det\left( \operatorname{sub}_{1,2,\ldots,i}^{1,2,\ldots ,i-1,j}A\right) . \end{equation} Define an $n \times n$-matrix $L_A =\left( \left( -1\right) ^{i+j}c_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq n}$ by \begin{equation} c_{i,j} = \begin{cases} \det\left( \operatorname{sub}_{1,2,\ldots,\widehat{j},\ldots,i}^{1,2,\ldots,i-1}A\right) , & \text{ if } j \leq i; \\ 0, & \text{ if } j > i . \end{cases} \end{equation} Here, the hat over the $j$ is a "magician's hat", which makes whatever comes under it disappear (so "$1,2,\ldots,\widehat{j},\ldots,i$" means "$1,2,\ldots,j-1,j+1,j+2,\ldots,i$").

Then:

Theorem 1. We have $R_A = L_A A$.

Proposition 2. The matrix $R_A$ is upper-triangular, while the matrix $L_A$ is lower-triangular.

We will prove these two facts below. Once they are proven, you can conclude that $A$ has the LU-decomposition $A = L_A^{-1} R_A$ when $L_A$ is invertible (since Proposition 2 shows that $L_A^{-1}$ is lower-triangular and $R_A$ is upper-triangular, but Theorem 1 yields $A = L_A^{-1} R_A$). When is $L_A$ invertible? The matrix $L_A$ is lower-triangular, so its invertibility is equivalent to the invertibility (in the base ring $\mathbb{K}$) of its diagonal entries. But its diagonal entries are $c_{i,i} = \det\left( \operatorname{sub}_{1,2,\ldots,\widehat{i},\ldots,i}^{1,2,\ldots,i-1}A\right) = \det\left( \operatorname{sub}_{1,2,\ldots,i-1}^{1,2,\ldots,i-1}A\right)$ for $i \in \left\{1,2,\ldots,n\right\}$, which are exactly the "proper northwestern principal minors" of $A$ (that is, all northwestern principal minors except for $\det A$ itself). Thus, you can conclude the following:

Corollary 3. If the "proper northwestern principal minors" of $A$ (that is, all the determinants $\det\left( \operatorname{sub}_{1,2,\ldots,i-1}^{1,2,\ldots,i-1}A\right)$ for $i \in \left\{1,2,\ldots,n\right\}$) are invertible (in the base ring $\mathbb{K}$), then $A$ has the LU-decomposition $A = L_A^{-1} R_A$.

Note that Corollary 3 is only a sufficient condition for the existence of an LU-decomposition. It is not a necessary one (as the example in which $n \geq 2$ and $A$ is the zero matrix shows: the zero matrix has an LU-decomposition, but for $n \geq 2$ it has a proper northwestern principal minor equal to $0$). But it is necessary if $A$ is invertible (since then, both the L and the U factors must be invertible, but this means that their diagonal entries are invertible; but the northwestern principal minors of $A$ are merely products of these diagonal entries).

Proofs

Proof of Theorem 1. Write the $n\times n$-matrix $A$ in the form $A=\left( a_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq n}$.

From $L_{A}=\left( \left( -1\right) ^{i+j}c_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq n}$ and $A=\left( a_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq n}$, we obtain \begin{equation} L_{A}A=\left( \sum_{k=1}^{n}\left( -1\right) ^{i+k}c_{i,k}a_{k,j}\right) _{1\leq i\leq n,\ 1\leq j\leq n}. \label{darij1.pf.t1.LAA=} \tag{1} \end{equation}

We must prove that $R_{A}=L_{A}A$. In view of \eqref{darij1.pf.t1.LAA=} and $R_{A}=\left( b_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq n}$, this boils down to proving that \begin{equation} b_{i,j}=\sum_{k=1}^{n}\left( -1\right) ^{i+k}c_{i,k}a_{k,j} \label{darij1.pf.t1.goal} \tag{2} \end{equation} for each $i\in\left\{ 1,2,\ldots,n\right\} $ and $j\in\left\{ 1,2,\ldots,n\right\} $. So let us prove \eqref{darij1.pf.t1.goal}.

Fix $i\in\left\{ 1,2,\ldots,n\right\} $ and $j\in\left\{ 1,2,\ldots ,n\right\} $. Consider the $i\times i$-matrix \begin{equation} \operatorname{sub}_{1,2,\ldots,i}^{1,2,\ldots,i-1,j}A=\left( \begin{array} [c]{ccccc} a_{1,1} & a_{1,2} & \cdots & a_{1,i-1} & a_{1,j}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,i-1} & a_{2,j}\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ a_{i-1,1} & a_{i-1,2} & \cdots & a_{i-1,i-1} & a_{i-1,j}\\ a_{i,1} & a_{i,2} & \cdots & a_{i,i-1} & a_{i,j} \end{array} \right) . \end{equation} Expanding the determinant of this matrix along its $i$-th column (whose entries are the first $i$ entries $a_{1,j},a_{2,j},\ldots,a_{i,j}$ of the $j$-th column of $A$) yields \begin{align*} & \det\left( \operatorname{sub}_{1,2,\ldots,i}^{1,2,\ldots,i-1,j}A\right) \\ & =\sum_{k=1}^{i}\left( -1\right) ^{i+k}a_{k,j}\det\left( \operatorname{sub}_{1,2,\ldots,\widehat{k},\ldots,i}^{1,2,\ldots ,i-1}A\right) \end{align*} (indeed, if $k\in\left\{ 1,2,\ldots,i\right\} $, then removing the $k$-th row and the $i$-th column from the matrix $\operatorname{sub}_{1,2,\ldots,i}^{1,2,\ldots,i-1,j}A$ yields the matrix $\operatorname{sub} _{1,2,\ldots,\widehat{k},\ldots,i}^{1,2,\ldots,i-1}A$). Comparing this with \begin{align*} & \sum_{k=1}^{n}\left( -1\right) ^{i+k}c_{i,k}a_{k,j}\\ & =\sum_{k=1}^{i}\left( -1\right) ^{i+k}\underbrace{c_{i,k}} _{\substack{=\det\left( \operatorname{sub}_{1,2,\ldots,\widehat{k},\ldots ,i}^{1,2,\ldots,i-1}A\right) \\\text{(by the definition of }c_{i,k} \text{,}\\\text{since }k\leq i\text{)}}}a_{k,j}+\sum_{k=i+1}^{n}\left( -1\right) ^{i+k}\underbrace{c_{i,k}}_{\substack{=0\\\text{(by the definition of }c_{i,k}\text{,}\\\text{since }k>i\text{)}}}a_{k,j}\\ & =\sum_{k=1}^{i}\left( -1\right) ^{i+k}\det\left( \operatorname{sub} _{1,2,\ldots,\widehat{k},\ldots,i}^{1,2,\ldots,i-1}A\right) a_{k,j} +\underbrace{\sum_{k=i+1}^{n}\left( -1\right) ^{i+k}0a_{k,j}}_{=0}\\ & =\sum_{k=1}^{i}\left( -1\right) ^{i+k}\det\left( \operatorname{sub} _{1,2,\ldots,\widehat{k},\ldots,i}^{1,2,\ldots,i-1}A\right) a_{k,j}\\ & =\sum_{k=1}^{i}\left( -1\right) ^{i+k}a_{k,j}\det\left( \operatorname{sub}_{1,2,\ldots,\widehat{k},\ldots,i}^{1,2,\ldots ,i-1}A\right) , \end{align*} we obtain \begin{equation} \sum_{k=1}^{n}\left( -1\right) ^{i+k}c_{i,k}a_{k,j}=\det\left( \operatorname{sub}_{1,2,\ldots,i}^{1,2,\ldots,i-1,j}A\right) =b_{i,j} \end{equation} (by the definition of $b_{i,j}$). This proves \eqref{darij1.pf.t1.goal}.

Thus, $R_{A}=L_{A}A$ holds. This proves Theorem 1. $\blacksquare$

Proof of Proposition 2. For any $i\in\left\{ 1,2,\ldots,n\right\} $ and $j\in\left\{ 1,2,\ldots,n\right\} $ satisfying $j>i$, we have $c_{i,j}=0$ (by the definition of $c_{i,j}$) and thus $\left( -1\right) ^{i+j} \underbrace{c_{i,j}}_{=0}=0$. Hence, the matrix $\left( \left( -1\right) ^{i+j}c_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq n}$ is lower-triangular. In other words, the matrix $L_{A}$ is lower-triangular (since $L_{A}=\left( \left( -1\right) ^{i+j}c_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq n}$).

It remains to prove that the matrix $R_{A}$ is upper-triangular. Indeed, let $i\in\left\{ 1,2,\ldots,n\right\} $ and $j\in\left\{ 1,2,\ldots,n\right\} $ be such that $i>j$. Hence, $j<i$, so that $j\in\left\{ 1,2,\ldots ,i-1\right\} $. Thus, the matrix $\operatorname{sub}_{1,2,\ldots ,i}^{1,2,\ldots,i-1,j}A$ has two equal columns (namely, its $j$-th column equals its $i$-th column). Thus, its determinant is $0$. In other words, $\det\left( \operatorname{sub}_{1,2,\ldots,i}^{1,2,\ldots,i-1,j}A\right) =0$. The definition of $b_{i,j}$ yields $b_{i,j}=\det\left( \operatorname{sub}_{1,2,\ldots,i}^{1,2,\ldots,i-1,j}A\right) =0$.

Now, forget that we fixed $i$ and $j$. We thus have shown that $b_{i,j}=0$ for all $i\in\left\{ 1,2,\ldots,n\right\} $ and $j\in\left\{ 1,2,\ldots ,n\right\} $ satisfying $i>j$. In other words, the matrix $\left( b_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq n}$ is upper-triangular. In other words, the matrix $R_{A}$ is upper-triangular. This completes the proof of Proposition 2. $\blacksquare$