矩阵LU分解为何不唯一？是否存在标准形式或允许多种形式？

Great question! Let's break this down clearly, starting with why LU decompositions aren't unique, then moving to standard forms that can fix this uniqueness issue.

First, let's clarify the core definition: an LU decomposition writes a matrix ( A ) as ( A = L U ), where ( L ) is lower-triangular and ( U ) is upper-triangular. The lack of uniqueness comes down to scaling freedom with diagonal matrices.

Here's the key idea: suppose we have a valid LU decomposition ( A = L U ). Let ( D ) be any invertible diagonal matrix (meaning all its diagonal entries are non-zero). We can rewrite the decomposition as:

A = (L D) * (D⁻¹ U)

• ( L D ) is still lower-triangular: multiplying a lower-triangular matrix by a diagonal matrix preserves the lower-triangular structure (non-diagonal entries stay the same, diagonal entries get scaled by ( D )'s elements).
• ( D⁻¹ U ) is still upper-triangular: same logic—diagonal matrix inverse is diagonal, multiplying by upper-triangular keeps it upper-triangular.

Since ( D ) can be any invertible diagonal matrix (not just the identity), this gives us infinitely many valid LU decompositions for most matrices.

Concrete Example #

Take ( A = \begin{bmatrix} 2 & 4 \ 1 & 3 \end{bmatrix} ):

• One decomposition: ( L = \begin{bmatrix} 2 & 0 \ 1 & 1 \end{bmatrix} ), ( U = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} )
• Pick ( D = \begin{bmatrix} 3 & 0 \ 0 & 2 \end{bmatrix} ), then:
  • ( L' = L D = \begin{bmatrix} 6 & 0 \ 3 & 2 \end{bmatrix} )
  • ( U' = D⁻¹ U = \begin{bmatrix} 1/3 & 2/3 \ 0 & 1/2 \end{bmatrix} )
• Multiply ( L' U' ), and you'll get back ( A )—this is a completely different valid decomposition!

While general LU decompositions aren't unique, we can add constraints to force uniqueness. These are the "standard forms" you're asking about:

1. Unit Lower-Triangular LU Decomposition #

If we require ( L ) to be a unit lower-triangular matrix (all diagonal entries equal to 1), then the decomposition becomes unique—provided all leading principal minors of ( A ) are non-zero (this is the key condition for existence and uniqueness here).

Why no more scaling freedom? Because if we tried to use the diagonal matrix trick above, ( L D ) would have diagonal entries equal to ( D )'s elements, which would no longer be 1. This violates the unit lower-triangular constraint, so we can't modify the decomposition without breaking the rules.

2. Unit Upper-Triangular LU Decomposition #

Alternatively, we can fix ( U ) to be a unit upper-triangular matrix (diagonal entries = 1). Just like the above, this also forces a unique decomposition (again, when leading principal minors are non-zero).

3. LDU Decomposition #

A more structured standard form is the LDU decomposition, where we split the decomposition into three parts:

A = L D U

• ( L ): unit lower-triangular
• ( D ): diagonal matrix
• ( U ): unit upper-triangular

This decomposition is unique (when leading principal minors are non-zero) because ( D )'s diagonal entries are uniquely determined by the ratios of ( A )'s leading principal minors, and ( L ) and ( U ) are fixed to be unit triangular (no scaling room left).

If ( A ) has a zero leading principal minor, an LU decomposition might not exist at all (without using a permutation matrix, i.e., PLU decomposition). Even if it does exist, uniqueness can fail even with unit triangular constraints—but those are special cases.

内容的提问来源于stack exchange，提问作者Mohammad Riazi-Kermani