# LU Factorization/Decomposition

I have the following matrix:

$A = \begin{bmatrix} 1 & 2 & -3 \\ -5 & -4 & 13 \\ 2 & 1 & -5 \\ \end{bmatrix}$

If you attempt to find the LU using what my book calls "storage of multipliers" or I assume to really be pivoting you run into a problem, when using this technique. That is U cannot have a value that is 0 in the diagonal entries.

In which case the following should be the wrong answer?

$L = \begin{bmatrix} 1 & 0 & 0 \\ -5 & 1 & 0 \\ 2 & -1/2 & 1 \\ \end{bmatrix}$

$U = \begin{bmatrix} 1 & 2 & -3 \\ 0 & 6 & -2 \\ 0 & 0 & 0 \\ \end{bmatrix}$

In which case one would have to use a permutation matrix P? (This wasn't covered for us but I assume it's the only way to solve the problem correctly, since U is not of the correct form?)

I'm basically confused as to if my answer is wrong or it's wrong and there is a more appropriate way to solve the problem. Thanks.

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 The use of a $P$ matrix is usually referred as partial pivoting and is needed to improve numerical conditioning. – user13838 Oct 23 '11 at 18:58

## 1 Answer

Why do you say that's the "wrong answer"?

If you're worried about the 0 on the diagonal, there's nothing you can do about it: since $A$ is singular, $\det A=0$ and so $\det L \det U = 0$ since determinants are multiplicative. Since the determinant of a triangular matrix is just the product of the diagonal entries, this means at least one of $L$ and $U$ must have a zero on the diagonal. $L$ is constrained to have all 1s on the diagonal, so it must be $U$.

What you may be remembering is that not every singular matrix has a LU decomposition. But when they do (such as this A) it is expected for some of the entries of $U$ to be zero.

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 Ah okay thanks for the explanation that was where I was confused was the 0 in the diagonal. – eWizardII Oct 23 '11 at 20:32