675: Complementary Subspace of Vectors Subspace

<The previous article in this series | The table of contents of this series | The next article in this series>

definition of complementary subspace of vectors subspace

---

Topics

About: vectors space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Note 1
• 4: Note 2
• 5: Note 3

---

Starting Context

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

---

Target Context

• The reader will have a definition of complementary subspace of vectors subspace.

---

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:
$F$: $\in \{\text{ the fields }\}$
$V^{\prime }$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$V$: $\in \{\text{ the vectors subspaces of }{V}^{\prime }\}$
$\ast \tilde{V}$: $\in \{\text{ the vectors subspaces of }{V}^{\prime }\}$
//

Conditions:
$V\cap \tilde{V}=\{0\}$
$\wedge$
$V^{\prime }=V+\tilde{V}$
//

$V^{\prime }=V+\tilde{V}$ means that $\mathrm{\forall }{v}^{\prime }\in V^{\prime }(\mathrm{\exists }v\in V,\mathrm{\exists }\tilde{v}\in \tilde{V}(v^{\prime }=v+\tilde{v}))$.

---

2: Natural Language Description

For any field, $F$, any $F$ vectors space, $V^{\prime }$, and any subspace, $V\subseteq V^{\prime }$, any subspace, $\tilde{V}\subseteq V^{\prime }$, such that $V\cap \tilde{V}=\{0\}$ and $V^{\prime }=V+\tilde{V}$

---

3: Note 1

The decomposition, $v^{\prime }=v+\tilde{v}$, is inevitably unique, because if $v^{\prime }=v_1+\widetilde{v_1}=v_2+\widetilde{v_2}$, $v_1-v_2=\widetilde{v_2}-\widetilde{v_1}$, but the left hand side is an element of $V$ and the right hand side is an element of $\tilde{V}$, so, it is in $V\cap \tilde{V}=\{0\}$, so, $v_1-v_2=\widetilde{v_2}-\widetilde{v_1}=0$, so, $v_1=v_2$ and $\widetilde{v_1}=\widetilde{v_2}$.

---

4: Note 2

There can be multiple complementary subspaces for the same vectors subspace, because for example, $V^{\prime }$ is the span of the basis, $\{e_1,e_2\}$, $V$ is the span of the basis, $\{e_1\}$, and the span of the basis, $\{e_2\}$, is a complementary subspace of $V$, but also the span of the basis, $\{e_1+e_2\}$, is a complementary subspace of $V$.

---

5: Note 3

$\tilde{V}$ does not need to be "perpendicular" to $V$ in order for it to be a complementary subspace of $V$. In fact, "perpendicular" is defined only when inner product is defined on $V^{\prime }$, and inner product is not required for the concept of complementary subspace.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>