To prove that if matrix $U$ is obtained from $A$ by the row operation $R_i = R_i + kR_j$, then $L$ is obtained from the identity matrix $I_n$ by the reverse row operation $R_i = R_i - kR_j$, such that $A = LU$, we can express $A$ as the product of $L$ and $U$, and then show that the given row operations on $A$ and $I_n$ lead to the formation of $L$ and $U$ satisfying $A = LU$.
Is this statement true? If yes how can I show this?
My approcah:
Given the LU decomposition $A = LU$, where: \begin{itemize} \item $L$ is a lower triangular matrix with ones on the diagonal. \item $U$ is an upper triangular matrix. \end{itemize}
We express $A$ as the product of $L$ and $U$: [ A = \begin{pmatrix} l_{11} & 0 & \cdots & 0 \\ l_{21} & l_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ l_{n1} & l_{n2} & \cdots & l_{nn} \end{pmatrix} \begin{pmatrix} u_{11} & u_{12} & \cdots & u_{1n} \\ 0 & u_{22} & \cdots & u_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & u_{nn} \end{pmatrix} ]
To find $L$ and $U$, we'll compare elements of $A$ with their corresponding elements in $L$ and $U$:
\begin{enumerate} \item From the definition of matrix multiplication, we find that the element $a_{ij}$ of $A$ is given by the dot product of the $i$-th row of $L$ and the $j$-th column of $U$: [ a_{ij} = l_{i1}u_{1j} + l_{i2}u_{2j} + \cdots + l_{in}u_{nj} ]
\item Using the given structure of $L$ and $U$, we observe that $a_{ij}$ can also be expressed as:
\[ a_{ij} = \sum_{k=1}^{n} l_{ik}u_{kj} \]
\item Equating the two expressions for $a_{ij}$, we have:
\[ a_{ij} = l_{i1}u_{1j} + l_{i2}u_{2j} + \cdots + l_{in}u_{nj} = \sum_{k=1}^{n} l_{ik}u_{kj} \]
\item This equation holds for all $i$ and $j$, implying that each element of $A$ is equal to the corresponding element of the matrix product $LU$.
\item Since $A = LU$, the given row operations on $A$ and $I_n$ will indeed lead to the formation of $L$ and $U$ matrices satisfying $A = LU$.
\end{enumerate}