description/proof of that for \(n \times n\) ring matrix, iff matrix multiplied by each column or row is original column or row, matrix is identity
Topics
About: ring
The table of contents of this article
Starting Context
- The reader knows a definition of ring of \(n \times n\) ring matrices.
Target Context
- The reader will have a description and a proof of the proposition that for any \(n \times n\) ring matrix, if and only if the matrix multiplied by each column or row is the original column or row, the matrix is the identity.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in M_n (R)\)
//
Statements:
(
\(\forall m \in R^m (M m = m)\), where \(m\) is regarded to be the column
\(\iff\)
\(M = I\)
)
\(\land\)
(
\(\forall m \in R^m (m M = m)\), where \(m\) is regarded to be the row
\(\iff\)
\(M = I\)
)
//
2: Proof
Whole Strategy: Step 1: suppose that \(\forall m \in R^m (M m = m)\); Step 2: see that \(M = I\); Step 3: suppose that \(M = I\); Step 4: see that \(\forall m \in R^m (M m = m)\); Step 5: suppose that \(\forall m \in R^m (m M = m)\); Step 6: see that \(M = I\); Step 7: suppose that \(M = I\); Step 8: see that \(\forall m \in R^m (m M = m)\).
Step 1:
Let us suppose that \(\forall m \in R^m (M m = m)\).
Step 2:
\(M^j_l m^l = m^j\).
Let \(p \in \{1, ..., n\}\) be any.
Let us take \(m\) such that \(m^p = 1\) and \(m^j = 0\) for each \(j \neq p\).
\(M^p_l m^l = m^p = 1\), but the left hand side is \(M^p_p m^p = M^p_p 1 = M^p_p\), so, \(M^p_p = 1\).
For each \(j \neq p\), \(M^j_l m^l = m^j = 0\), but the left hand side is \(M^j_p m^p = M^j_p 1 = M^j_p\), so, \(M^j_p = 0\).
So, each diagonal component of \(M\) is \(1\) and the other components of \(M\) are \(0\).
So, \(M = I\).
Step 3:
Let us suppose that \(M = I\).
Step 4:
For each \(m \in R^m\), \((M m)^j = M^j_l m^l = I^j_l m^l = m^j\).
So, \(M m = m\).
Step 5:
Let us suppose that \(\forall m \in R^m (m M = m)\).
Step 6:
\(m_l M^l_j = m_j\).
Let \(p \in \{1, ..., n\}\) be any.
Let us take \(m\) such that \(m_p = 1\) and \(m_j = 0\) for each \(j \neq p\).
\(m_l M^l_p = m_p = 1\), but the left hand side is \(m_p M^p_p = 1 M^p_p = M^p_p\), so, \(M^p_p = 1\).
For each \(j \neq p\), \(m_l M^l_j = m_j = 0\), but the left hand side is \(m_p M^p_j = 1 M^p_j = M^p_j\), so, \(M^p_j = 0\).
So, each diagonal component of \(M\) is \(1\) and the other components of \(M\) are \(0\).
So, \(M = I\).
Step 7:
Let us suppose that \(M = I\).
Step 8:
For each \(m \in R^m\), \((m M)_j = m_l M^l_j = m_l I^l_j = m_j\).
So, \(m M = m\).
