Frobenius Inequality

Let $\mathbf{B}$ be an $m \times r$ matrix of rank $r$. Then there exists a matrix $\mathbf{X}$ (called a left inverse of B), such that $\mathbf{X B}=\mathbf{I}$.

PROOF. If $m=r$, then $\mathbf{B}$ is nonsingular and admits an inverse. So suppose $r<m$. The columns of $\mathbf{B}$ are linearly independent. Thus we can find a set of $m-r$ columns that together with the columns of $\mathbf{B}$ form a basis for $R^{m}$. In other words, we can find a matrix $\mathbf{U}$ of order $m \times(m-r)$ such that $[\mathbf{B}, \mathbf{U}]$ is nonsingular. Let the inverse of $[\mathbf{B}, \mathbf{U}]$ be partitioned as $\left[\begin{array}{l}\mathbf{X} \ \mathbf{V}\end{array}\right]$, where $\mathbf{X}$ is $r \times m$. Since
$$
\left[\begin{array}{l}
\mathbf{X} \
\mathbf{V}
\end{array}\right][\mathbf{B}, \mathbf{U}]=\mathbf{I}
$$
we have $\mathbf{X B}=\mathbf{I}$
We can similarly show that an $r \times n$ matrix $\mathbf{C}$ of rank $r$ has a right inverse, i.e., a matrix $\mathbf{Y}$ such that $\mathbf{C Y}=\mathbf{I}$. Note that a left inverse or a right inverse is not unique, unless the matrix is square and nonsingular.

We can similarly show that an $r \times n$ matrix $\mathbf{C}$ of rank $r$ has a right inverse, i.e., a matrix $\mathbf{Y}$ such that $\mathbf{C Y}=\mathbf{I}$. Note that a left inverse or a rig

Let $\mathbf{B}$ be an $m \times r$ matrix of rank $r$. Then there exists a nonsingular matrix P such that
$$
\mathbf{P B}=\left[\begin{array}{l}
\mathbf{I} \
\mathbf{0}
\end{array}\right]
$$

ht inverse is not unique, unless the matrix is square and nonsingular.

Similarly, if $\mathbf{C}$ is $r \times n$ of rank $r$, then there exists a nonsingular matrix $\mathbf{Q}$ such that $\mathbf{C Q}=[\mathbf{I}, \mathbf{0}]$. These two results and the rank factorization (see $\mathbf{4 . 3})$ immediately lead to the following.

发表回复

您的电子邮箱地址不会被公开。