738: For Vectors Space, Subspace, and Complementary Subspace, Finite-Dimensional Subspace That Intersects Complementary Subspace Trivially Is Projected into Subspace as Same-Dimensional Subspace

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description/proof of that for vectors space, subspace, and complementary subspace, finite-dimensional subspace that intersects complementary subspace trivially is projected into subspace as same-dimensional subspace

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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: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of projection of vector into vectors subspace with respect to complementary subspace.
• The reader admits the proposition that any projection from any vectors space into any subspace with respect to any complementary subspace is a linear map, and the image of any subspace under the projection is a subspace.
• The reader admits the proposition that for any linear map from any finite-dimensional vectors space, there is a domain subspace that is 'vectors spaces - linear morphisms' isomorphic to the map range by the restriction of the map on the subspace domain.
• The reader admits the proposition that any 2 finite dimensional vectors spaces such that there is a linear bijection from one of them to the other are of the same dimension..

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Target Context

• The reader will have a description and a proof of the proposition that for any vectors space, any subspace, and any complementary subspace, any finite-dimensional subspace that intersects the complementary subspace trivially is projected into the subspace as a same-dimensional subspace.

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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:
$F$: $\in \{\text{ the fields }\}$
$V^{\prime }$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$V$: $\in \{\text{ the subspaces of }{V}^{\prime }\}$
$\tilde{V}$: $\in \{\text{ the complementary subspaces of }V\}$
$f$: $V^{\prime }\to V$, $=\text{ the projection into }V\text{ with respect to }\tilde{V}$
$V^{″}$: $\in \{\text{ the }d\text{ -dimensional subspaces of }{V}^{\prime }\}$
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Statements:
$V^{″}\cap \tilde{V}=\{0\}$
$⟹$
$f(V^{″})\in \{\text{ the }d\text{ -dimensional subspaces of }{V}^{\prime }\}$
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2: Natural Language Description

For any field, $F$, any $F$ vectors space, $V^{\prime }$, any subspace, $V\subseteq V^{\prime }$, any complementary subspace, $\tilde{V}\subseteq V^{\prime }$, the projection into $V$ with respect to $\tilde{V}$, $f:V^{\prime }\to V$, and any d-dimensional subspace, $V^{″}\subseteq V^{\prime }$, such that $V^{″}\cap \tilde{V}=\{0\}$, $f(V^{″})$ is a d-dimensional subspace of $V^{\prime }$.

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3: Proof

Whole Strategy: Step 1: see that $f(V^{″})$ is a subspace of $V^{\prime }$; Step 2: choose a subspace, $V^{‴}\subseteq V^{″}$, that is 'vectors spaces - linear morphisms' isomorphic to $f(V^{″})$; Step 3: suppose that the dimension of $V^{‴}$ was smaller than $d$ and find a contradiction; Step 4: see that the dimension of $f(V^{″})$ is $d$.

Step 1:

$f(V^{″})$ is a subspace of $V^{\prime }$, by the proposition that any projection from any vectors space into any subspace with respect to any complementary subspace is a linear map, and the image of any subspace under the projection is a subspace.

Step 2:

There is a subspace, $V^{‴}\subseteq V^{″}$, that is 'vectors spaces - linear morphisms' isomorphic to $f(V^{″})$ by the restriction of $f$ on the $V^{‴}$ domain, $f^{\prime }:V^{‴}\to f(V^{″})$, by the proposition that for any linear map from any finite-dimensional vectors space, there is a domain subspace that is 'vectors spaces - linear morphisms' isomorphic to the map range by the restriction of the map on the subspace domain.

Step 3:

Let us suppose the dimension of $V^{‴}$ to be $l$. Also the dimension of $f(V^{″})$ is $l$, by the proposition that any 2 finite dimensional vectors spaces such that there is a linear bijection from one of them to the other are of the same dimension..

Let us suppose that $l<d$.

There would be a basis for $V^{‴}$, $\{e_1,...,e_l\}$. There would be an element, $v\in V^{″}$, that was not any linear combination of the basis. $\{f^{\prime }(e_1),...,f^{\prime }(e_l)\}$ would be a basis for $f(V^{″})$, because $f^{\prime }$ was a 'vectors spaces - linear morphisms' isomorphism. $f(v)=c^1{f}^{\prime }(e_1)+...+c^l{f}^{\prime }(e_l)$. Let us define $v^{\prime }=v-(c^1{e}_1+...+c^l{e}_l)$, nonzero, because $v$ was not any linear combination of the basis. $f(v^{\prime })=f(v-(c^1{e}_1+...+c^l{e}_l))=f(v)-(c^{1}f(e_1)+...+c^{l}f(e_l))$, because $f$ was linear, by the proposition that any projection from any vectors space into any subspace with respect to any complementary subspace is a linear map, and the image of any subspace under the projection is a subspace, $=f(v)-(c^1{f}^{\prime }(e_1)+...+c^l{f}^{\prime }(e_l))=0$. Then, $v^{\prime }=0+v^{″}=v^{″}$ for a nonzero $v^{″}\in \tilde{V}$, which would imply that $V^{″}$ intersected $\tilde{V}$ non-trivially, a contradiction.

So, $l=d$.

Step 4:

As the dimension of $f(V^{″})$ is $l$, it is $l=d$.

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References

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