description/proof of that for (n + n') x (n + n'') injective matrix with right-top n x n'' submatrix 0, matrix with left-top n x n submatrix replaced with injective matrix is injective
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of injection.
- The reader knows a definition of %field name% matrices space.
Target Context
- The reader will have a description and a proof of the proposition that for any (n + n') x (n + n'') injective matrix with the right-top n x n'' submatrix 0, the matrix with the left-top n x n submatrix replaced with any injective matrix is injective.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(N\): \(\in \{\text{ the } (n + n') x (n + n'') F \text{ injective matrices } \}\), with the top-right \(n x n''\) matrix 0
\(M\): \(\in \{\text{ the } n x n F \text{ injective matrices } \}\)
\(\tilde{N}\): \(= N \text{ with the top-left } n x n \text{ submatrix replaced with } M\), \(\in \{\text{ the } (n + n') x (n + n'') F \text{ matrices } \}\)
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Statements:
\(\tilde{N} \in \{\text{ the injective matrices }\}\)
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"injective matrix" means that the canonical linear map induced by the matrix is injective.
2: Proof
Whole Strategy: Step 1: decompose any vector, \(w \in F^{n + n''}\), as \(w = v + v''\), where the \(n + 1 \sim n + n'\) components of \(v\) are \(0\) and the \(1 \sim n\) components of \(v''\) are \(0\); Step 2: take any \(\tilde{w} = \tilde{v} + \tilde{v''}\) such that \(w \neq \tilde{w}\), and see that \(v \neq \tilde{v}\) or \(v = \tilde{v} \land v'' \neq \tilde{v''}\); Step 3: see that when \(v \neq \tilde{v}\), \(\tilde{N} w \neq \tilde{N} \tilde{w}\); Step 4: see that when \(v = \tilde{v} \land v'' \neq \tilde{v''}\), \(\tilde{N} v'' \neq \tilde{N} \tilde{v''}\) and so, \(\tilde{N} w \neq \tilde{N} \tilde{w}\); Step 5: conclude the proposition.
Step 1:
Let us decompose any vector, \(w \in F^{n + n''}\), as \(w = v + v''\), where the \(n + 1 \sim n + n'\) components of \(v\) are \(0\) and the \(1 \sim n\) components of \(v''\) are \(0\).
Step 2:
Let \(\tilde{w} \in F^{n + n''}\) be any such that \(w \neq \tilde{w}\).
\(\tilde{w}\) is decomposed to be \(= \tilde{v} + \tilde{v''}\).
\(v \neq \tilde{v}\) or \(v = \tilde{v}\).
When \(v = \tilde{v}\), \(v'' \neq \tilde{v''}\), because otherwise, it would be that \(w = v + v'' = \tilde{v} + \tilde{v''} = \tilde{w}\), a contradiction.
Step 3:
Let us suppose that \(v \neq \tilde{v}\).
For \(\tilde{N} w = \tilde{N} (v + v'') = \tilde{N} v + \tilde{N} v''\), the \(1 \sim n\) components are contributed only by \(\tilde{N} v\), because the \(1 \sim n\) rows of \(\tilde{N}\) have the \(0\) \(n + 1 \sim n + n''\) components and \(v''\) has the \(0\) \(1 \sim n\) components.
Likewise for \(\tilde{N} \tilde{w}\).
Denoting the projection of \(F^{n + n''}\) into the 1st \(F^n\) as \(\pi_n\), the \(1 \sim n\) components of \(\tilde{N} v\) are \(M \pi_n (v)\) and the \(1 \sim n\) components of \(\tilde{N} \tilde{v}\) are \(M \pi_n (\tilde{v})\).
\(M \pi_n (v) \neq M \pi_n (\tilde{v})\), because \(M\) is injective.
So, \(\tilde{N} w \neq \tilde{N} \tilde{w}\).
Step 4:
Let us suppose that \(v = \tilde{v} \land v'' \neq \tilde{v''}\).
We already know that the \(1 \sim n\) components of \(\tilde{N} v''\) and the \(1 \sim n\) components of \(\tilde{N} \tilde{v''}\) are \(0\).
In fact, also the \(1 \sim n\) components of \(N v''\) and the \(1 \sim n\) components of \(N \tilde{v''}\) are \(0\), by the same reason.
The \(n + 1 \sim n + n'\) components of \(\tilde{N} v''\) equal the \(n + 1 \sim n + n'\) components of \(N v''\), because they do not depend on the replaced \(n x n\) submatrix.
Likewise, the \(n + 1 \sim n + n'\) components of \(\tilde{N} \tilde{v''}\) equal the \(n + 1 \sim n + n'\) components of \(N \tilde{v''}\).
So, \(N v'' = \tilde{N} v''\) and \(N \tilde{v''} = \tilde{N} \tilde{v''}\).
But as \(N\) is injective, \(N v'' \neq N \tilde{v''}\), and so, \(\tilde{N} v'' = N v'' \neq N \tilde{v''} = \tilde{N} \tilde{v''}\).
So, \(\tilde{N} w = \tilde{N} v + \tilde{N} v'' \neq \tilde{N} \tilde{v} + \tilde{N} \tilde{v''} = \tilde{N} \tilde{w}\).
Step 5:
So, \(\tilde{N} w \neq \tilde{N} \tilde{w}\) anyway.
So, \(\tilde{N}\) is injective.
