Exercise 2.3.8

Question

Show that if the equation $A \mathbf{x} = \mathbf{0}$ has unique solution (i.e. if echelon form of $A$ has pivot in every column), then $A$ is left invertible.

Jing Huang · Accepted Answer

\begin{proof}
$A \mathbf{x} = \mathbf{0}$ has unique solution, then the solution is trivial solution. The echelon form of A has pivot at every column. $A$ is $m 	imes n$ matrix, then $m \geq n$. The row number is greater or equal to the column number. The reduced echelon form of $A$ is denoted as
%%%%%%%%%%%%%%%%%
\begin{equation*}
    A_{re} = 
    \begin{bmatrix}
    I_{n 	imes n} \
    \mathbf{0}_{(m - n) 	imes n} \
    \end{bmatrix}.
\end{equation*}
%%%%%%%%%%%%%%%
And suppose $A_{re}$ is obtained by a sequence of elementary row operation $E_1, E_2, ... , E_k$,
%%%%%%%%%%%%%%%%%
\begin{equation*}
    A_{re} = E_k \; ... \; E_2 E_1 A
\end{equation*}
%%%%%%%%%%%%%%%%
$E_i$ is $m 	imes m$. The left inverse of $A$ is the first n rows of the product of $E_i$. i.e.
%%%%%%%%%%%%%%%%%
\begin{equation*}
    E_{left} = I_{n 	imes m} E_k \; ... \; E_2 E_1,
\end{equation*}
%%%%%%%%%%%%%%%%
where
%%%%%%%%%%%%%%%%%
\begin{equation*}
    I_{n 	imes m} = 
    \begin{bmatrix}
    1   & 0     & 0     &\dots  & 0 \
    0   & 1      & 0    &\dots  & 0 \
    \vdots & \vdots & \vdots & \ddots & \vdots \
    0   & 0      & \dots    &1   & \dots \
    \end{bmatrix}_{n 	imes m},
\end{equation*}
%%%%%%%%%%%%%%%%
is used to extract the $I_{n 	imes n}$ identity matrix in $A_{re}$. $E_{left}A = I$ so $A$ is left invertible.
\end{proof}

Exercise 2.3.8

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Exercise 2.3.8

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