Finding the A=LDU factorization for matrices

Question

I am trying to factorize the matrix on the picture, but I am not sure how can I get the upper triangular matrix. Do I have to reverse every step of the matrix operation that I performed? Any help will be greatly appreciated!

Answer 1

I usually work out row reduction in Gauss-Jordan form: \begin{align} A=\begin{bmatrix} 1&0&1\\ 2&2&2\\ 3&4&5 \end{bmatrix} &\xrightarrow{\substack{E_{31}(-3)\\E_{21}(-2)}} \begin{bmatrix} 1&0&1\\ 0&2&0\\ 0&4&2 \end{bmatrix} \\[6px]&\xrightarrow{E_2(1/2)} \begin{bmatrix} 1&0&1\\ 0&1&0\\ 0&4&2 \end{bmatrix} \\[6px]&\xrightarrow{E_{32}(-4)} \begin{bmatrix} 1&0&1\\ 0&1&0\\ 0&0&2 \end{bmatrix} \\[6px]&\xrightarrow{E_{3}(1/2)} \begin{bmatrix} 1&0&1\\ 0&1&0\\ 0&0&1 \end{bmatrix}=U \end{align} The notation used is

• $E_i(c)$ means multiplying the $i$th row by $c\ne0$
• $E_{ij}(d)$ means summing to the $i$th row the $j$th row multiplied by $d$

These can be seen as elementary matrices that successively multiply on the left, so I get $$ L_0=E_{21}(2)E_{31}(3)E_2(2)E_{31}(4)E_3(2)= \begin{bmatrix} 1 & 0 & 0 \\ 2 & 2 & 0 \\ 3 & 4 & 2 \end{bmatrix} $$ the product of the inverses in the reverse order.

It can be shown that the entries are exactly as dictated by the elementary matrices, with $0$ added off-diagonal and missing entries on the diagonal are $1$.

Now writing $L_0=LD$ is easy: $$ L= \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix} \qquad D=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} $$ (divide each column by the corresponding entry in the diagonal of $D$).

Answer 2

I am including the $L$ component of the answer as well. But $U$ is explained clearly too:

The goal of row reduction is to reduce a matrix $A$ to an upper-triangular matrix $A^{'}$, which you have as a step in your picture. $$A=\begin{bmatrix}1&0&1\\2&2&2\\3&4&5\end{bmatrix}\to \, \, A^{'} = \begin{bmatrix}1&0&1\\0&2&0\\0&0&2\end{bmatrix} $$

But each row operation can be represented as multiplying $A$on the left by some matrix $E$. In your picture, you use 3 row operations to get to $A^{'}$, so $ E = E_1 E_2 E_3 $ and we have $$ EA = A^{'}$$

Now we can solve for $A$ and see $$A = E^{-1} A^{'} $$

Note that $A^{'}$ can be written as the product of a diagonal matrix and an upper triangular matrix with 1 on the diagonals, $A^{'} = DU = \begin{bmatrix}1&0&0\\0&2&0\\0&0&2\end{bmatrix} \begin{bmatrix}1&0&1\\0&1&0\\0&0&1\end{bmatrix}$.

It turns out that $E^{-1}$ is already lower triangular, so we have

$$ A = LDU = E^{-1} \begin{bmatrix}1&0&0\\0&2&0\\0&0&2\end{bmatrix} \begin{bmatrix}1&0&1\\0&1&0\\0&0&1\end{bmatrix} $$