1093: For Vectors Space and 2 Subspaces, if Sum of Projections into Subspaces Is Projection into Subspace, 2 Subspaces Are Perpendicular to Each Other w.r.t. Projections

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description/proof of that for vectors space and 2 subspaces, if sum of projections into subspaces is projection into subspace, 2 subspaces are perpendicular to each other w.r.t. projections

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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: Note
• 3: Proof

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Starting Context

• The reader knows a definition of projection from vectors space into vectors subspace.

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Target Context

• The reader will have a description and a proof of the proposition that for any vectors space and any 2 subspaces, if the sum of any projections into the subspaces is any projection into any another subspace, the 2 subspaces are perpendicular to each other with respect to the projections.

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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:
$F$: $\in \{\text{ the fields }\}$
$V^{\prime }$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$V_1$: $\in \{\text{ the vectors subspaces of }{V}^{\prime }\}$
$V_2$: $\in \{\text{ the vectors subspaces of }{V}^{\prime }\}$
$V_3$: $\in \{\text{ the vectors subspaces of }{V}^{\prime }\}$
$f_1$: $:V^{\prime }\to V_1$, $\in \{\text{ the projections }\}$
$f_2$: $:V^{\prime }\to V_2$, $\in \{\text{ the projections }\}$
$f_3$: $:V^{\prime }\to V_3$, $\in \{\text{ the projections }\}$
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Statements:
$f_1+f_2=f_3$
$⟹$
$\mathrm{\forall }{v}_1\in V_1(f_2(v_1)=0)\wedge \mathrm{\forall }{v}_2\in V_2(f_1(v_2)=0)$
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2: Note

$f_1$, $f_2$, and $f_3$ are defined somewhat independently according to the definition of projection from vectors space into vectors subspace, but as far as they satisfies the definition and $f_1+f_2=f_3$ is satisfied, this proposition holds as Proof shows.

Typically, $f_1$, $f_2$, and $f_3$ are defined uniformly according to the definition of orthogonal projection from vectors space with inner product into vectors subspace, and this proposition holds as a special case.

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3: Proof

Whole Strategy: Step 1: do $f_3\circ (f_1+f_2)(v_1)=f_3\circ f_3(v_1)$ and see that $f_1\circ f_2(v_1)+f_2(v_1)=0$; Step 2: see that $V_1\cap V_2=\{0\}$; Step 3: see that $f_2(v_1)=0$; Step 4: conclude the proposition.

Step 1:

As $f_1+f_2=f_3$, $f_3\circ (f_1+f_2)=f_3\circ f_3=f_3$.

Let $v_1\in V_1$ be any.

$f_3\circ (f_1+f_2)(v_1)=f_3(v_1)$.

$f_3\circ (f_1+f_2)(v_1)=f_3(f_1(v_1)+f_2(v_1))=f_3(v_1+f_2(v_1))=f_3(v_1)+f_3\circ f_2(v_1)$.

So, $f_3(v_1)+f_3\circ f_2(v_1)=f_3(v_1)$, which implies that $f_3\circ f_2(v_1)=0$.

$(f_1+f_2)\circ f_2(v_1)=0$, so, $f_1\circ f_2(v_1)+f_2\circ f_2(v_1)=f_1\circ f_2(v_1)+f_2(v_1)=0$.

Step 2:

Let us see that $V_1\cap V_2=\{0\}$.

Let $v^{\prime }\in V_1\cap V_2$ be any.

As $f_1+f_2=f_3$, $(f_1+f_2)(v^{\prime })=f_3(v^{\prime })$.

But $(f_1+f_2)(v^{\prime })=f_1(v^{\prime })+f_2(v^{\prime })=v^{\prime }+v^{\prime }=2v^{\prime }$.

So, $f_3(v^{\prime })=2v^{\prime }$.

$f_3\circ f_3(v^{\prime })=f_3(2v^{\prime })$, which implies that $f_3(v^{\prime })=2f_3(v^{\prime })$, so, $0=f_3(v^{\prime })-f_3(v^{\prime })=2f_3(v^{\prime })-f_3(v^{\prime })=f_3(v^{\prime })$. But as $f_3(v^{\prime })=2v^{\prime }$, $2v^{\prime }=0$, so, $v^{\prime }=0/2=0$.

That means that $V_1\cap V_2=\{0\}$.

Step 3:

By Step 1, $f_1\circ f_2(v_1)=-f_2(v_1)$.

But the left hand side is in $V_1$ and the right hand side is in $V_2$, so, the both hand sides are in $V_1\cap V_2$.

But by Step 2, $V_1\cap V_2=\{0\}$, and $-f_2(v_1)=0$, which implies that $f_2(v_1)=0$.

Step 4:

By symmetry, for each $v_2\in V_2$, $f_1(v_2)=0$.

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References

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