# LU decomposition steps

I've been looking at some LU Decomposition problems and I understand that making a matrix A reduced to the form A=LU , where L is a lower triangular matrix and U is a upper triangular matrix, however I am having trouble understanding the steps to get to these matrices. Could someone please explain the method for LU Decomposition in detail, preferably excluding the concept of permutation matrices? ( We haven't talked about permutation matrices in class yet so our professor forbids us to use them for the decomposition).

## 1 Answer

$LU$ decomposition is really just another way to say Gaussian elimination. If you're familiar with that, putting the pieces together is easy. Here is an example. Let $$A=A^{\left(0\right)}=\left[\begin{array}{ccc} 8 & 1 & 6\\ 4 & 9 & 2\\ 0 & 5 & 7 \end{array}\right].$$ Proceed by Gaussian elimination. The first multiplier is $\ell_{2,1}=4/8=0.5$ (this is the multiplier that allows us to cancel $a_{2,1}=4$ using the first row) and the second is $\ell_{3,1}=0/8=0$. We arrive at $$A^{\left(1\right)}=\left[\begin{array}{ccc} 8 & 1 & 6\\ 0 & 8.5 & -1\\ 0 & 5 & 7 \end{array}\right].$$ To cancel out $a_{3,2}^{\left(1\right)}=5$, we use the multiplier $\ell_{3,2}=5/8.5\approx0.5882$ to yield $$A^{\left(2\right)}\approx\left[\begin{array}{ccc} 8 & 1 & 6\\ 0 & 8.5 & -1\\ 0 & 0 & 7.5882 \end{array}\right]$$ which yields the $LU$ decomposition $$A=LU\approx\left[\begin{array}{ccc} 1 & 0 & 0\\ 0.5 & 1 & 0\\ 0 & 0.5882 & 1 \end{array}\right]\left[\begin{array}{ccc} 8 & 1 & 6\\ 0 & 8.5 & -1\\ 0 & 0 & 7.5882 \end{array}\right].$$ Note that $L$ is just made up of the multipliers we used in Gaussian elimination with $1$s on the diagonal, while $U$ is just $A^{\left(2\right)}$.

• Very interesting. Thank you for explaining it so well Commented Apr 10, 2019 at 15:08