847: 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

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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

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of injection.
• The reader knows a definition of %field name% matrices space.

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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.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\text{ the fields }\}$
$N$: $\in \{\text{ the }(n+n^{\prime })x(n+n^{″})F\text{ injective matrices }\}$, with the top-right $nx{n}^{″}$ matrix 0
$M$: $\in \{\text{ the }nxnF\text{ injective matrices }\}$
$\tilde{N}$: $=N\text{ with the top-left }nxn\text{ submatrix replaced with }M$, $\in \{\text{ the }(n+n^{\prime })x(n+n^{″})F\text{ matrices }\}$
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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.

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2: Proof

Whole Strategy: Step 1: decompose any vector, $w\in F^{n+n^{″}}$, as $w=v+v^{″}$, where the $n+1\sim n+n^{\prime }$ components of $v$ are $0$ and the $1\sim n$ components of $v^{″}$ are $0$; Step 2: take any $\tilde{w}=\tilde{v}+\widetilde{v^{″}}$ such that $w\ne \tilde{w}$, and see that $v\ne \tilde{v}$ or $v=\tilde{v}\wedge v^{″}\ne \widetilde{v^{″}}$; Step 3: see that when $v\ne \tilde{v}$, $\tilde{N}w\ne \tilde{N}\tilde{w}$; Step 4: see that when $v=\tilde{v}\wedge v^{″}\ne \widetilde{v^{″}}$, $\tilde{N}{v}^{″}\ne \tilde{N}\widetilde{v^{″}}$ and so, $\tilde{N}w\ne \tilde{N}\tilde{w}$; Step 5: conclude the proposition.

Step 1:

Let us decompose any vector, $w\in F^{n+n^{″}}$, as $w=v+v^{″}$, where the $n+1\sim n+n^{\prime }$ 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 $w\ne \tilde{w}$.

$\tilde{w}$ is decomposed to be $=\tilde{v}+\widetilde{v^{″}}$.

$v\ne \tilde{v}$ or $v=\tilde{v}$.

When $v=\tilde{v}$, $v^{″}\ne \widetilde{v^{″}}$, because otherwise, it would be that $w=v+v^{″}=\tilde{v}+\widetilde{v^{″}}=\tilde{w}$, a contradiction.

Step 3:

Let us suppose that $v\ne \tilde{v}$.

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 $M{\pi }_n(\tilde{v})$.

$M{\pi }_n(v)\ne M{\pi }_n(\tilde{v})$, because $M$ is injective.

So, $\tilde{N}w\ne \tilde{N}\tilde{w}$.

Step 4:

Let us suppose that $v=\tilde{v}\wedge v^{″}\ne \widetilde{v^{″}}$.

We already know that the $1\sim n$ components of $\tilde{N}{v}^{″}$ and the $1\sim n$ components of $\tilde{N}\widetilde{v^{″}}$ are $0$.

In fact, also the $1\sim n$ components of $N{v}^{″}$ and the $1\sim n$ components of $N\widetilde{v^{″}}$ are $0$, by the same reason.

The $n+1\sim n+n^{\prime }$ components of $\tilde{N}{v}^{″}$ equal the $n+1\sim n+n^{\prime }$ components of $N{v}^{″}$, because they do not depend on the replaced $nxn$ submatrix.

Likewise, the $n+1\sim n+n^{\prime }$ components of $\tilde{N}\widetilde{v^{″}}$ equal the $n+1\sim n+n^{\prime }$ components of $N\widetilde{v^{″}}$.

So, $N{v}^{″}=\tilde{N}{v}^{″}$ and $N\widetilde{v^{″}}=\tilde{N}\widetilde{v^{″}}$.

But as $N$ is injective, $N{v}^{″}\ne N\widetilde{v^{″}}$, and so, $\tilde{N}{v}^{″}=N{v}^{″}\ne N\widetilde{v^{″}}=\tilde{N}\widetilde{v^{″}}$.

So, $\tilde{N}w=\tilde{N}v+\tilde{N}{v}^{″}\ne \tilde{N}\tilde{v}+\tilde{N}\widetilde{v^{″}}=\tilde{N}\tilde{w}$.

Step 5:

So, $\tilde{N}w\ne \tilde{N}\tilde{w}$ anyway.

So, $\tilde{N}$ is injective.

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References

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