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
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of projection from vectors space into vectors subspace.
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.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V'\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V_1\): \(\in \{\text{ the vectors subspaces of } V'\}\)
\(V_2\): \(\in \{\text{ the vectors subspaces of } V'\}\)
\(V_3\): \(\in \{\text{ the vectors subspaces of } V'\}\)
\(f_1\): \(: V' \to V_1\), \(\in \{\text{ the projections }\}\)
\(f_2\): \(: V' \to V_2\), \(\in \{\text{ the projections }\}\)
\(f_3\): \(: V' \to V_3\), \(\in \{\text{ the projections }\}\)
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Statements:
\(f_1 + f_2 = f_3\)
\(\implies\)
\(\forall v_1 \in V_1 (f_2 (v_1) = 0) \land \forall v_2 \in V_2 (f_1 (v_2) = 0)\)
//
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.
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 \(v' \in V_1 \cap V_2\) be any.
As \(f_1 + f_2 = f_3\), \((f_1 + f_2) (v') = f_3 (v')\).
But \((f_1 + f_2) (v') = f_1 (v') + f_2 (v') = v' + v' = 2 v'\).
So, \(f_3 (v') = 2 v'\).
\(f_3 \circ f_3 (v') = f_3 (2 v')\), which implies that \(f_3 (v') = 2 f_3 (v')\), so, \(0 = f_3 (v') - f_3 (v') = 2 f_3 (v') - f_3 (v') = f_3 (v')\). But as \(f_3 (v') = 2 v'\), \(2 v' = 0\), so, \(v' = 0 / 2 = 0\).
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 \(v_2 \in V_2\), \(f_1 (v_2) = 0\).
