1315: For Lie Group Homomorphism, if Differential at Identity Is 0, Map Is Constant over Connected Component

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

description/proof of that for Lie group homomorphism, if differential at identity Is $0$, map is constant over connected component

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Topics

About: $C^{\mathrm{\infty }}$ manifold

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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 Lie group.
• The reader knows a definition of %structure kind name% homomorphism.
• The reader knows a definition of differential of $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds with boundary at point.
• The reader knows a definition of connected topological component.
• The reader admits the proposition that for any 2 Lie group homomorphisms between any Lie groups, the map as the multiplication of the 1st homomorphism and the multiplicative inverse of the 2nd homomorphism has a constant rank.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ map between any $C^{\mathrm{\infty }}$ manifolds, if the map has the constant rank $0$, the map is constant over each connected component.

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Target Context

• The reader will have a description and a proof of the proposition that for any Lie group homomorphism, if the differential at the identity is $0$, the map is constant over each connected component.

---

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:
$G_1$: $\in \{\text{ the Lie groups }\}$
$G_2$: $\in \{\text{ the Lie groups }\}$
$f$: $:G_1\to G_2$, $\in \{\text{ the Lie group homomorphisms }\}$
$d{f}_1$: $:T_1{G}_1\to T_1{G}_2$, $=\text{ the differential at }1$
//

Statements:
$d{f}_1=0$
$⟹$
$\mathrm{\forall }T\in \{\text{ the connected components of }{M}_1\}(f{|}_T\in \{\text{ the constant maps }\})$
//

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2: Note

Especially, for the connected component that contains $1$, $T_1$, $f{|}_T=1$, because $f{|}_T(1)=1$.

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3: Proof

Whole Strategy: Step 1: see that $f$ is a constant rank map; Step 2: conclude the proposition.

Step 1:

Any Lie group homomorphism has a constant rank, by the proposition that for any 2 Lie group homomorphisms between any Lie groups, the map as the multiplication of the 1st homomorphism and the multiplicative inverse of the 2nd homomorphism has a constant rank: the immediate corollary mentioned in the Note.

So, $f$ has a constant rank, but as $f$ has the rank, $0$, at $1$, $f$ has the constant $0$ rank.

Step 2:

Let $T$ be any connected component of $M_1$.

By the proposition that for any $C^{\mathrm{\infty }}$ map between any $C^{\mathrm{\infty }}$ manifolds, if the map has the constant rank $0$, the map is constant over each connected component, $f{|}_T$ is constant.

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References

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

1314: For 2 Lie Group Homomorphisms Between Lie Groups, Map as Multiplication of 1st Homomorphism and Multiplicative Inverse of 2nd Homomorphism Has Constant Rank

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

description/proof of that for 2 Lie group homomorphisms between Lie groups, map as multiplication of 1st homomorphism and multiplicative inverse of 2nd homomorphism has constant rank

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Topics

About: $C^{\mathrm{\infty }}$ manifold

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of Lie group.
• The reader knows a definition of %structure kind name% homomorphism.
• The reader knows a definition of rank of $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds with boundary at point.
• The reader admits the proposition that for any linear map between any vectors spaces, any 'vectors spaces - linear morphisms' isomorphism onto the domain of the linear map, and any 'vectors spaces - linear morphisms' isomorphism from any superspace of the codomain of the linear map, the composition of the linear map after the 1st isomorphism and before the 2nd isomorphism has the rank of the linear map.

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Target Context

• The reader will have a description and a proof of the proposition that for any 2 Lie group homomorphisms between any Lie groups, the map as the multiplication of the 1st homomorphism and the multiplicative inverse of the 2nd homomorphism has a constant rank.

---

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:
$G_1$: $\in \{\text{ the Lie groups }\}$
$G_2$: $\in \{\text{ the Lie groups }\}$
$f$: $:G_1\to G_2$, $\in \{\text{ the Lie group homomorphisms }\}$
$f^{\prime }$: $:G_1\to G_2$, $\in \{\text{ the Lie group homomorphisms }\}$
$h$: $:G_1\to G_2,g\mapsto f(g)(f^{\prime }(g){)}^{-1}$
//

Statements:
$\mathrm{\forall }g\in G_1(Rank({dh}_g)=Rank({dh}_1))$
//

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2: Note

$h$ is not any Lie group homomorphism in general, because $h(g{g}^{\prime })=f(g{g}^{\prime })(f^{\prime }(g{g}^{\prime }){)}^{-1}=f(g)f(g^{\prime })(f^{\prime }(g)f^{\prime }(g^{\prime }){)}^{-1}=f(g)f(g^{\prime })(f^{\prime }(g^{\prime }){)}^{-1}(f^{\prime }(g){)}^{-1}=f(g)h(g^{\prime })(f^{\prime }(g){)}^{-1}$, which is not guaranteed to equal $h(g)h(g^{\prime })=f(g)(f^{\prime }(g){)}^{-1}f(g^{\prime })(f^{\prime }(g^{\prime }){)}^{-1}$, unless $G_2$ is Abelian.

As an immediate corollary, $f$ has a constant rank, because $f^{\prime }$ can be taken to be $f^{\prime }=1$, which is a Lie group homomorphism, and $h(g)=f(g)(f^{\prime }(g){)}^{-1}=f(g)1^{-1}=f(g)1=f(g)$, so, $h=f$.

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3: Proof

Whole Strategy: Step 1: see that $h(g_{0}g)=R_{(f^{\prime }(g_0){)}^{-1}}\circ L_{f(g_0)}\circ h(g)$ and take $h^{\prime }:G_1\to G_2,g\mapsto h(g_{0}g),=R_{(f^{\prime }(g_0){)}^{-1}}\circ L_{f(g_0)}\circ h=h\circ L_{g_0}$; Step 2: see that ${d{h}^{\prime }}_1={d{R}_{(f^{\prime }(g_0){)}^{-1}}}_{f(g_0)}\circ {d{L}_{f(g_0)}}_1\circ {dh}_1={dh}_{g_0}\circ {d{L}_{g_0}}_1$; Step 3: conclude the proposition.

Step 1:

$h$ is $C^{\mathrm{\infty }}$, because $f$ and $f^{\prime }$ are $C^{\mathrm{\infty }}$, and the multiplicative inverse map and the multiplication map are $C^{\mathrm{\infty }}$: it is $:G_1\to G_1\times G_1\to G_2\times G_2\to G_2\times G_2\to G_2,g\mapsto (g,g)\mapsto (f(g),f^{\prime }(g))\mapsto (f(g),(f^{\prime }(g){)}^{-1})\mapsto f(g)(f^{\prime }(g){)}^{-1}$.

For each $g_0\in G_j$, let $L_{g_0}:G_j\to G_j,g\mapsto g_{0}g$ and $R_{g_0}:G_j\to G_j,g\mapsto g{g}_0$ be the left translation and the right translation by $g_0$, which are some diffeomorphisms.

For each $g_0,g\in G_1$, let us see that $h(g_{0}g)=R_{(f^{\prime }(g_0){)}^{-1}}\circ L_{f(g_0)}\circ h(g)$.

$h(g_{0}g)=f(g_{0}g)(f^{\prime }(g_{0}g){)}^{-1}=f(g_0)f(g)(f^{\prime }(g_0)f^{\prime }(g){)}^{-1}=f(g_0)f(g)(f^{\prime }(g){)}^{-1}(f^{\prime }(g_0){)}^{-1}=f(g_0)h(g)(f^{\prime }(g_0){)}^{-1}=L_{f(g_0)}(h(g))(f^{\prime }(g_0){)}^{-1}=R_{(f^{\prime }(g_0){)}^{-1}}(L_{f(g_0)}(h(g)))$.

So, let us take $h^{\prime }:G_1\to G_2,g\mapsto h(g_{0}g),=R_{(f^{\prime }(g_0){)}^{-1}}\circ L_{f(g_0)}\circ h=h\circ L_{g_0}$, which is $C^{\mathrm{\infty }}$ as a composition of $C^{\mathrm{\infty }}$ maps.

Step 2:

${d{h}^{\prime }}_1={d{R}_{(f^{\prime }(g_0){)}^{-1}}}_{f(g_0)h(1)}\circ {d{L}_{f(g_0)}}_{h(1)}\circ {dh}_1={dh}_{g_{0}1}\circ {d{L}_{g_0}}_1$.

As $h(1)=f(1)(f^{\prime }(1){)}^{-1}=11^{-1}=11=1$, ${d{h}^{\prime }}_1={d{R}_{(f^{\prime }(g_0){)}^{-1}}}_{f(g_0)}\circ {d{L}_{f(g_0)}}_1\circ {dh}_1={dh}_{g_0}\circ {d{L}_{g_0}}_1$.

Step 3:

${d{R}_{(f^{\prime }(g_0){)}^{-1}}}_{f(g_0)}:T_{f(g_0)}{G}_2\to T_{f(g_0)(f^{\prime }(g_0){)}^{-1}}{G}_2$, which is a 'vectors spaces - linear morphisms' isomorphism, because $R_{(f^{\prime }(g_0){)}^{-1}}$ is a diffeomorphism.

${d{L}_{f(g_0)}}_1:T_1{G}_2\to T_{f(g_0)}{G}_2$, which is a 'vectors spaces - linear morphisms' isomorphism, because $L_{f(g_0)}$ is a diffeomorphism.

${dh}_1:T_1{G}_1\to T_1{G}_2$.

${dh}_{g_0}:T_{g_0}{G}_1\to T_{h(g_0)}{G}_2$.

${d{L}_{g_0}}_1:T_1{G}_1\to T_{g_0}{G}_1$, which is a 'vectors spaces - linear morphisms' isomorphism, because $L_{g_0}$ is a diffeomorphism.

By the proposition that for any linear map between any vectors spaces, any 'vectors spaces - linear morphisms' isomorphism onto the domain of the linear map, and any 'vectors spaces - linear morphisms' isomorphism from any superspace of the codomain of the linear map, the composition of the linear map after the 1st isomorphism and before the 2nd isomorphism has the rank of the linear map, $Rank({d{h}^{\prime }}_1)=Rank({dh}_1)=Rank({dh}_{g_0})$.

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References

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

1313: For C∞ Map Between C∞ Manifolds, if Map Has Constant Rank 0, Map Is Constant over Connected Component

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

description/proof of that for $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds, if map has constant rank $0$, map is constant over connected component

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Topics

About: $C^{\mathrm{\infty }}$ manifold

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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

---

Starting Context

• The reader knows a definition of rank of $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds with boundary at point.
• The reader knows a definition of connected topological component.
• The reader admits the rank theorem for $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds.
• The reader admits the proposition that any connected topological component is exactly any connected topological subspace that cannot be made larger.
• The reader admits the proposition that any map that is anywhere locally constant on any connected topological space is globally constant.

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Target Context

• The reader will have a description and a proof of the proposition that for any $C^{\mathrm{\infty }}$ map between any $C^{\mathrm{\infty }}$ manifolds, if the map has the constant rank $0$, the map is constant over each connected component.

---

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:
$M_1$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds }\}$
$M_2$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds }\}$
$f$: $:M_1\to M_2$, $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
//

Statements:
$\mathrm{\forall }m\in M_1(Rank(f{)}_m=0)$
$⟹$
$\mathrm{\forall }T\in \{\text{ the connected components of }{M}_1\}(f{|}_T\in \{\text{ the constant maps }\})$
//

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2: Note

$M_1$ and $M_2$ are required to be without boundary for this proposition, because this proposition uses the rank theorem for $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds, which requires the domain and the codomain without boundary.

As an immediate corollary, when $M_1$ is connected, $f$ is constant.

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3: Proof

Whole Strategy: Step 1: see that around each $m\in M_1$, there is an open neighborhood, $U_m$, over which $f$ is constant; Step 2: conclude the proposition.

Step 1:

Let $m\in M_1$ be any.

As $f$ has the constant rank, $0$, by the rank theorem for $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds, there are a chart around $m$, $(U_m\subseteq M_1,{\varphi }_m)$, and a chart around $f(m)$, $(U_{f(m)}\subseteq M_2,{\varphi }_{f(m)})$, such that ${\varphi }_{f(m)}\circ f\circ {{\varphi }_m}^{-1}{|}_{{\varphi }_m(U_m)}:{\varphi }_m(U_m)\to {\varphi }_{f(m)}(U_{f(m)})$ is $(x^1,...,x^{d_1})\mapsto (0,...,0)$.

That means that $f$ is constant over $U_m$.

Step 2:

Let $T\subseteq M_1$ be any connected component of $M_1$.

$T$ is a connected topological subspace of $M_1$, by the proposition that any connected topological component is exactly any connected topological subspace that cannot be made larger.

$f{|}_T:T\to M_2$ is a map that is anywhere locally constant over a connected topological space: around each $t\in T$, there is an open neighborhood of $t$ on $M_1$, $U_t\subseteq M_1$, over which $f$ is constant by Step 1, and $U_t\cap T\subseteq T$ is an open neighborhood of $t$ on $T$ over which $f{|}_T$ is constant.

By the proposition that any map that is anywhere locally constant on any connected topological space is globally constant, $f{|}_T$ is globally constant.

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References

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

1312: Rank of C∞ Map Between C∞ Manifolds with Boundary at Point

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

definition of rank of $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds with boundary at point

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Topics

About: $C^{\mathrm{\infty }}$ manifold

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of differential of $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds with boundary at point.
• The reader knows a definition of rank of linear map between vectors spaces.

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Target Context

• The reader will have a definition of rank of $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds with boundary at point.

---

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:
$M_1$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$M_2$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$f$: $:M_1\to M_2$, $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
$m$: $\in M_1$
$d{f}_m$: $:T_m{M}_1\to T_{f(m)}{M}_2$, $=\text{ the differential of }f\text{ at }m$
$\ast Rank(f{)}_m$: $=Rank(d{f}_m)$
//

Conditions:
//

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2: Note

In general, the rank of $f$ changes point to point.

When the rank of $f$ is the same at all the points of $M_1$, $f$ is called to have "constant rank".

When at each point of $M_1$, there is a neighborhood of the point over which $f$ has the same rank, $f$ is called to have "locally constant rank".

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References

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

1311: For Linear Map Between Vectors Spaces, 'Vectors Spaces - Linear Morphisms' Isomorphism onto Domain of Linear Map, and 'Vectors Spaces - Linear Morphisms' Isomorphism from Superspace of Codomain of Linear Map, Composition of Linear Map After 1st Isomorphism and Before 2nd Isomorphism Has Rank of Linear Map

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

description/proof of that for linear map between vectors spaces, 'vectors spaces - linear morphisms' isomorphism onto domain of linear map, and 'vectors spaces - linear morphisms' isomorphism from superspace of codomain of linear map, composition of linear map after 1st isomorphism and before 2nd isomorphism has rank of linear map

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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

---

Starting Context

• The reader knows a definition of %field name% vectors space.
• The reader knows a definition of %category name% isomorphism.
• The reader knows a definition of rank of linear map between vectors spaces.
• The reader admits the proposition that the range of any linear map between any vectors spaces is a vectors subspace of the codomain.
• The reader admits the proposition that for any 'vectors spaces - linear morphisms' isomorphism, its restriction on any subspace domain and the corresponding range codomain is a 'vectors spaces - linear morphisms' isomorphism.
• The reader admits the proposition that for any 'vectors spaces - linear morphisms' isomorphism, the image of any linearly independent subset or any basis of the domain is linearly independent or a basis on the codomain.

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Target Context

• The reader will have a description and a proof of the proposition that for any linear map between any vectors spaces, any 'vectors spaces - linear morphisms' isomorphism onto the domain of the linear map, and any 'vectors spaces - linear morphisms' isomorphism from any superspace of the codomain of the linear map, the composition of the linear map after the 1st isomorphism and before the 2nd isomorphism has the rank of the linear map.

---

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:
$F$: $\in \{\text{ the fields }\}$
$V_1$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$V_2$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$f_1$: $:V_1\to V_2$, $\in \{\text{ the linear maps }\}$
$V_0$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$f_0$: $:V_0\to V_1$, $\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}$
$V_2^{\prime }$: $\in \{\text{ the }F\text{ vectors spaces }\}$, such that $V_2\in \{\text{ the vectors subspaces of }{V}_2^{\prime }\}$
$V_3$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$f_2$: $:V_2^{\prime }\to V_3$, $\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}$
//

Statements:
$Rank(f_2\circ f_1\circ f_0)=Rank(f_1)$
//

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2: Note

The codomain of $f_0$ is $V_1$ with the vectors space structure of $V_1$ and $V_2$ is a vectors subspace of $V_2^{\prime }$ (not just a subset), which are the points.

Refer to the proposition that for any linear map between any vectors spaces and any 'vectors spaces - linear morphisms' isomorphism, the composition of the linear map after the isomorphism does not necessarily have the rank of the linear map.

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3: Proof

Whole Strategy: Step 1: see that $f_2\circ f_1\circ f_0$ is linear; Step 2: see that $Ran(f_1\circ f_0)=Ran(f_1)$; Step 3: see that $f_2{|}_{Ran(f_1)}:Ran(f_1)\to f_2(Ran(f_1))$ is a 'vectors spaces - linear morphisms' isomorphism, see that $f_2\circ f_1\circ f_0=f_2{|}_{Ran(f_1)}\circ f_1\circ f_0$, and see that $Dim(Ran(f_2\circ f_1\circ f_0))=Dim(Ran(f_1\circ f_0))$.

Step 1:

$f_2\circ f_1\circ f_0$ is linear, which is because the codomain of $f_0$ is the domain of $f_1$ (more generally, a vectors subspace of the domain of $f_1$) and the codomain of $f_1$ is a vectors subspace of the domain of $f_2$: just $V_1\subseteq V_1$ and $V_2\subseteq V_2^{\prime }$ sets-wise does not guarantee the linearity.

We needed to check the linearity, because otherwise, $Rank(f_2\circ f_1\circ f_0)$ would not be defined.

Step 2:

Let us see that $Ran(f_1\circ f_0)=Ran(f_1)$.

For each $v\in Ran(f_1\circ f_0)$, $v=f_1(f_0(v_0))$ for a $v_0\in V_0$, but $v_1:=f_0(v_0)\in V_1$ and $v=f_1(v_1)\in Ran(f_1)$.

For each $v\in Ran(f_1)$, $v=f_1(v_1)$ for a $v_1\in V_1$, but as $f_0$ is surjective, there is a $v_0\in V_0$ such that $v_1=f_0(v_0)$, so, $v=f_1(v_1)=f_1(f_0(v_0))\in Ran(f_1\circ f_0)$.

That implies that $Rank(f_1\circ f_0)=Dim(Ran(f_1\circ f_0))=Dim(Ran(f_1))=Rank(f_1)$.

Step 3:

$Ran(f_1)$ is a vectors subspace of $V_2$, by the proposition that the range of any linear map between any vectors spaces is a vectors subspace of the codomain.

$Ran(f_1)$ is a vectors subspace of $V_2^{\prime }$, because $V_2$ is a vectors subspace of $V_2^{\prime }$.

$f_2{|}_{Ran(f_1)}:Ran(f_1)\to f_2(Ran(f_1))$ is a 'vectors spaces - linear morphisms' isomorphism, by the proposition that for any 'vectors spaces - linear morphisms' isomorphism, its restriction on any subspace domain and the corresponding range codomain is a 'vectors spaces - linear morphisms' isomorphism.

$f_2\circ f_1\circ f_0=f_2{|}_{Ran(f_1)}\circ f_1\circ f_0$.

So, $Dim(Ran(f_2\circ f_1\circ f_0))=Dim(Ran(f_2{|}_{Ran(f_1)}\circ f_1\circ f_0))$.

But as $Ran(f_1\circ f_0)=Ran(f_1)$, $Ran(f_2{|}_{Ran(f_1)}\circ f_1\circ f_0)=Ran(f_2{|}_{Ran(f_1)})$, so, $Dim(Ran(f_2{|}_{Ran(f_1)}\circ f_1\circ f_0))=Dim(Ran(f_2{|}_{Ran(f_1)}))$, but $Dim(Ran(f_2{|}_{Ran(f_1)}))=Dim(Ran(f_1))$, by the proposition that for any 'vectors spaces - linear morphisms' isomorphism, the image of any linearly independent subset or any basis of the domain is linearly independent or a basis on the codomain, so, $Rank(f_2\circ f_1\circ f_0)=Dim(Ran(f_2\circ f_1\circ f_0))=Dim(Ran(f_2{|}_{Ran(f_1)}\circ f_1\circ f_0))=Dim(Ran(f_1))=Rank(f_1)$.

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References

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

1310: For Linear Map Between Vectors Spaces and 'Vectors Spaces - Linear Morphisms' Isomorphism, Composition of Linear Map after Isomorphism Does Not Necessarily Have Rank of Linear Map

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

description/proof of that for linear map between vectors spaces and 'vectors spaces - linear morphisms' isomorphism, composition of linear map after isomorphism does not necessarily have rank of linear map

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Topics

About: vectors space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of %field name% vectors space.
• The reader knows a definition of %category name% isomorphism.
• The reader knows a definition of rank of linear map between vectors spaces.
• The reader knows a definition of direct sum of modules.

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Target Context

• The reader will have a description and a proof of the proposition that for a linear map between some vectors spaces and a 'vectors spaces - linear morphisms' isomorphism, the composition of the linear map after the isomorphism does not necessarily have the rank of the linear map.

---

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:
$F$: $\in \{\text{ the fields }\}$
$V_1^{\prime }$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$V_2$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$f_1$: $:V_1^{\prime }\to V_2$, $\in \{\text{ the linear maps }\}$
$V_0$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$V_1$: $\in \{\text{ the }F\text{ vectors spaces }\}$ such that $V_1\subseteq V_1^{\prime }$
$f_0$: $:V_0\to V_1$, $\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}$
//

Statements:
Not necessarily $Rank(f_1\circ f_0)=Rank(f_1)$
//

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2: Note

$V_1$ is not presupposed to be a vectors subspace of $V_1^{\prime }$ and $V_1$ is not presupposed to equal $V_1^{\prime }$ (the composition is valid if $V_1\subseteq V_1^{\prime }$), which are the points.

This proposition is a warning not to jump to the conclusion without checking some things: refer to the proposition that for any linear map between any vectors spaces, any 'vectors spaces - linear morphisms' isomorphism onto the domain of the linear map, and any 'vectors spaces - linear morphisms' isomorphism from any vectors superspace of the codomain of the linear map, the composition of the linear map after the 1st isomorphism and before the 2nd isomorphism has the rank of the linear map.

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3: Proof

Whole Strategy: Step 1: see that $f_1\circ f_0$ is not guaranteed to be linear; Step 2: see a counterexample that even if $V_1$ is a vectors subspace of $V_1^{\prime }$, $Rank(f_1\circ f_0)<Rank(f_1)$.

Step 1:

$V_1\subseteq V_1^{\prime }$ means just that $V_1$ is a subset of $V_1^{\prime }$ not that $V_1$ is a vectors subspace of $V_1^{\prime }$.

For each $v,v^{\prime }\in V_0$ and each $r,r^{\prime }\in F$, $f_1\circ f_0(rv+r^{\prime }{v}^{\prime })=f_1(f_0(rv+r^{\prime }{v}^{\prime }))=f_1(r{f}_0(v)+r^{\prime }{f}_0(v^{\prime }))$, but the issue here is that $r{f}_0(v)+r^{\prime }{f}_0(v^{\prime })$ is by the vectors space structure of $V_1$ not of $V_1^{\prime }$, so, $=r{f}_1(f_0(v))+r^{\prime }{f}_1(f_0(v^{\prime }))$ is not guaranteed.

So, $f_1\circ f_0$ is not guaranteed to be linear.

Without being linear, $Rank(f_1\circ f_0)$ is not defined.

Step 2:

Let us suppose that $V_1$ is a vectors subspace of $V_1^{\prime }$.

Let $V_1^{\prime }={\mathbb{R}}^2$ as the Euclidean vectors space, $V_2={\mathbb{R}}^2$ as the Euclidean vectors space, $f_1=i{d}_{{\mathbb{R}}^2}:V_1^{\prime }\to V_2$ as the identity map, $V_0=\mathbb{R}\oplus \{0\}$ where $\mathbb{R}$ is the Euclidean vectors space, $V_1=\mathbb{R}\oplus \{0\}$ where $\mathbb{R}$ is the Euclidean vectors space, and $f_0=i{d}_{\mathbb{R}\oplus \{0\}}:V_0\to V_1$ as the identity map.

Certainly, $V_1\subseteq V_1^{\prime }$, where $V_1$ is 1-dimensional, because for example, $\{(1,0)\}$ is a basis.

Certainly, $f_0$ is a 'vectors spaces - linear morphisms' isomorphism.

But $Ran(f_1)={\mathbb{R}}^2$ and $Rank(f_1)=Dim(Ran(f_1))=2$, while $Ran(f_1\circ f_0)=\mathbb{R}\otimes \{0\}$ and $Rank(f_1\circ f_0)=Dim(Ran(f_1\circ f_0))=1$.

So, $Rank(f_1\circ f_0)<Rank(f_1)$.

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References

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

1309: For 'Vectors Spaces - Linear Morphisms' Isomorphism, Its Restriction on Subspace Domain and Corresponding Range Codomain Is 'Vectors Spaces - Linear Morphisms' Isomorphism

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description/proof of that for 'vectors spaces - linear morphisms' isomorphism, its restriction on subspace domain and corresponding range codomain is 'vectors spaces - linear morphisms' isomorphism

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Topics

About: vectors space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of %field name% vectors space.
• The reader knows a definition of %category name% isomorphism.
• The reader admits the proposition that the range of any linear map between any vectors spaces is a vectors subspace of the codomain.
• The reader admits the proposition that any bijective linear map between any vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.

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Target Context

• The reader will have a description and a proof of the proposition that for any 'vectors spaces - linear morphisms' isomorphism, its restriction on any subspace domain and the corresponding range codomain is a 'vectors spaces - linear morphisms' isomorphism.

---

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:
$F$: $\in \{\text{ the fields }\}$
$V_1^{\prime }$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$V_2^{\prime }$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$f^{\prime }$: $:V_1^{\prime }\to V_2^{\prime }$, $\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}$
$V_1$: $\in \{\text{ the vectors subspaces of }{V}_1^{\prime }\}$
$V_2$: $=f^{\prime }(V_1)$
$f$: $=f{|}_{V_1}:V_1\to V_2$
//

Statements:
$f\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}$
//

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2: Proof

Whole Strategy: Step 1: see that $V_2$ is a vectors subspace of $V_2^{\prime }$; Step 2: see that $f$ is linear; Step 3: see that $f$ is bijective; Step 4: conclude the proposition.

Step 1:

$V_2:=f^{\prime }(V_1)$ is a vectors subspace of $V_2^{\prime }$, by the proposition that the range of any linear map between any vectors spaces is a vectors subspace of the codomain, which means that $V_2$ is an $F$ vectors space.

So, $f$ is a map from an $F$ vectors space into an $F$ vectors space.

Step 2:

$f$ is linear, because for each $v_1,v_2\in V_1$ and each $r_1,r_2\in F$, $f(r_1{v}_1+r_2{v}_2)=f^{\prime }(r_1{v}_1+r_2{v}_2)=r_1{f}^{\prime }(v_1)+r_2{f}^{\prime }(v_2)=r_{1}f(v_1)+r_{2}f(v_2)$.

Step 3:

$f$ is bijective, because for each $v_1,v_2\in V_1$ such that $v_1\ne v_2$, $f(v_1)=f^{\prime }(v_1)\ne f^{\prime }(v_2)=f(v_2)$, because $f^{\prime }$ is injective, and $f$ is obviously surjective.

Step 4:

By the proposition that any bijective linear map between any vectors spaces is a 'vectors spaces - linear morphisms' isomorphism, $f$ is a 'vectors spaces - linear morphisms' isomorphism.

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References

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

1308: For Linear Map Between Finite-Dimensional Vectors Spaces, Rank of Map Is Rank of Representative Matrix w.r.t. Bases

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description/proof of that for linear map between finite-dimensional vectors spaces, rank of map is rank of representative matrix w.r.t. bases

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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

---

Starting Context

• The reader knows a definition of rank of linear map between vectors spaces.
• The reader knows a definition of rank of matrix over ring.
• The reader admits the proposition that for any finite-dimensional vectors space and any basis, the vectors space is 'vectors spaces - linear morphisms' isomorphic to the components vectors space with respect to the basis.
• The reader admits the proposition that for any 'vectors spaces - linear morphisms' isomorphism, the image of any linearly independent subset or any basis of the domain is linearly independent or a basis on the codomain.
• The reader admits the proposition that for any finite-dimensional columns or rows module and any linearly independent subset, any expansion of the subset into any larger-dimensional columns or rows module is linearly independent.
• The reader admits the proposition that for any finite-dimensional columns or rows vectors space and any linearly independent subset, the subset can be shrunk into a number-of-elements-dimensional columns or rows vectors space by choosing some components.
• The reader admits the proposition that for any vectors space, any finite generator can be reduced to be a basis.
• The reader admits the proposition that for any vectors space, the bases have the same cardinality.
• The reader admits the proposition that the determinant of any square matrix over any field is nonzero if and only if the set of the columns or the rows is linearly independent.

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Target Context

• The reader will have a description and a proof of the proposition that for any linear map between any finite-dimensional vectors spaces, the rank of the map is the rank of the representative matrix with respect to any bases.

---

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:
$F$: $\in \{\text{ the fields }\}$
$V_1$: $\in \{\text{ the }{d}_1\text{ -dimensional }F\text{ vectors spaces }\}$
$V_2$: $\in \{\text{ the }{d}_2\text{ -dimensional }F\text{ vectors spaces }\}$
$f$: $:V_1\to V_2$, $\in \{\text{ the linear maps }\}$
$B_1$: $\in \{\text{ the bases for }{V}_1\}$
$B_2$: $\in \{\text{ the bases for }{V}_2\}$
$M$: $=\text{ the representative matrix for }f$ with respect to $B_1$ and $B_2$
//

Statements:
$Rank(f)=Rank(M)$
//

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2: Proof

Whole Strategy: Step 1: see that $V_2$ is 'vectors spaces - linear morphisms' isomorphic to the components vectors space and see that the rank of $f$ is the rank of the linear map by $M$; Step 2: suppose that $Rank(M)=d$ and the top-left $d\times d$ submatrix is determinant nonzero; Step 3: see that $Rank(f)=d$; Step 4: see that when $Rank(f)=d$, $Rank(M)=d$.

Step 1:

$V_2$ is 'vectors spaces - linear morphisms' isomorphic to the components vectors space, by the proposition that for any finite-dimensional vectors space and any basis, the vectors space is 'vectors spaces - linear morphisms' isomorphic to the components vectors space with respect to the basis.

That means that any subspace of $V_2$ has the same dimension with the corresponding subspace of the components vectors space, by the proposition that for any 'vectors spaces - linear morphisms' isomorphism, the image of any linearly independent subset or any basis of the domain is linearly independent or a basis on the codomain: any basis of the subspace of $V_2$ is mapped to a basis of the corresponding subspace of the components vectors space: the restriction of the 'vectors spaces - linear morphisms' isomorphism is obviously a 'vectors spaces - linear morphisms' isomorphism.

So, $Rank(f)$ is the dimension of the range of the components vectors spaces linear map by $M$.

Step 2:

Let us suppose that $Rank(M)=d$.

Let the top-left $d\times d$ submatrix, $N$, be determinant nonzero, which is possible by reordering $B_1$ and $B_2$.

Step 3:

Let us take the set of $d$-dimensional column vectors, $\{(1,0,...,0{)}^t,...,(0,...,0,1{)}^t\}$, which is obviously linearly independent.

$\{N(1,0,...,0{)}^t,...,N(0,...,0,1{)}^t\}$, which is a set of $d$-dimensional column vectors, is linearly independent, by the proposition that for any 'vectors spaces - linear morphisms' isomorphism, the image of any linearly independent subset or any basis of the domain is linearly independent or a basis on the codomain: the linear map by $N$ is a 'vectors spaces - linear morphisms' isomorphism, because $detN\ne 0$.

Let us think of the $d_1$-dimensional expansion of $\{(1,0,...,0{)}^t,...,(0,...,0,1{)}^t\}$, $\{(1,0,...,0,0,...,0{)}^t,...,(0,...,0,1,0,...,0{)}^t\}$ and $\{M(1,0,...,0,0,...,0{)}^t,...,M(0,...,0,1,0,...,0{)}^t\}$.

$\{M(1,0,...,0,0,...,0{)}^t,...,M(0,...,0,1,0,...,0{)}^t\}$ is a $d_2$-dimensional expansion of $\{N(1,0,...,0{)}^t,...,N(0,...,0,1{)}^t\}$, because the 1st $d$ components are not changed, and so, is linearly independent, by the proposition that for any finite-dimensional columns or rows module and any linearly independent subset, any expansion of the subset into any larger-dimensional columns or rows module is linearly independent.

So, the range of the map by $M$ is equal to or larger than $d$-dimensional.

Let us suppose that the range was $d^{\prime }$-dimensional where $d<d^{\prime }$.

There would be a basis for the range, and some $d^{\prime }$ components could be chosen such that the shrunk basis was still linearly independent, by the proposition that for any finite-dimensional columns or rows vectors space and any linearly independent subset, the subset can be shrunk into a number-of-elements-dimensional columns or rows vectors space by choosing some components.

Let us move the chosen $d^{\prime }$ components to the top $d^{\prime }$ rows by reordering $B_2$, calling the result $M$ hereafter.

That would mean that the shrunk range (the projection to the 1st $d^{\prime }$ components) was $d^{\prime }$-dimensional.

Let $M^{\prime }$ be the top $d^{\prime }\times d_1$ submatrix of $M$.

$\{M^{\prime }\left(\begin{array}{c}1\\ 0\\ ...\\ 0\end{array}\right),...,M^{\prime }\left(\begin{array}{c}0\\ ...\\ 0\\ 1\end{array}\right)\}$ would span (generate) the shrunk range, obviously.

But it would be nothing but the columns of $M^{\prime }$.

So, some $d^{\prime }$ columns of $M^{\prime }$ could be chosen to be a basis for the shrunk range, by the proposition that for any vectors space, any finite generator can be reduced to be a basis and the proposition that for any vectors space, the bases have the same cardinality.

Let us move the corresponding $d^{\prime }$ columns of $M$ to the left by reordering $B_1$, calling the result $M$ hereafter.

Then, the top-left $d^{\prime }\times d^{\prime }$ submatrix would be determinant nonzero, by the proposition that the determinant of any square matrix over any field is nonzero if and only if the set of the columns or the rows is linearly independent, so, $d<d^{\prime }\le Rank(M)$, a contradiction.

So, the range is $d$-dimensional.

So, the rank of the linear map by $M$ is $d$, so, $Rank(f)=d$.

Step 4:

Let us suppose that $Rank(f)=d$.

If $Rank(M)=d^{\prime }$, $Rank(f)=d^{\prime }$, by Step 3, so, $d=d^{\prime }$.

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References

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

1307: Determinant of Square Matrix over Field Is Nonzero iff Set of Columns or Rows Is Linearly Independent

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

description/proof of that determinant of square matrix over field is nonzero iff set of columns or rows is linearly independent

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Topics

About: matrices space

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The table of contents of this article

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

---

Starting Context

• The reader knows a definition of field.
• The reader knows a definition of determinant of square matrix over ring.
• The reader knows a definition of linearly independent subset of module.
• The reader knows Cramer's rule for any system of linear equations.

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Target Context

• The reader will have a description and a proof of the proposition that the determinant of any square matrix over any field is nonzero if and only if the set of the columns or the rows is linearly independent.

---

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:
$F$: $\in \{\text{ the fields }\}$
$M$: $\in \{\text{ the }n\times nF\text{ matrices }\}$
//

Statements:
(
$detM\ne 0$
$⟺$
$\{\text{ the columns of }M\}\in \{\text{ the linearly independent subsets of the }n\text{ -dimensional }F\text{ columns vectors space }\}$
)
$\wedge$
(
$detM\ne 0$
$⟺$
$\{\text{ the rows of }M\}\in \{\text{ the linearly independent subsets of the }n\text{ -dimensional }F\text{ rows vectors space }\}$
)
//

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2: Proof

Whole Strategy: apply Cramer's rule for any system of linear equations; Step 1: take $M(c^1,...,c^n{)}^t=0$ and conclude for the columns; Step 2: take $M^t(c^1,...,c^n{)}^t=0$ and conclude for the rows.

Step 1:

Let the $j$-th column of $M$ be denoted as $M_j$.

Let us take $M(c^1,...,c^n{)}^t=0$.

It is nothing but $c^1{M}_1+...+c^n{M}_n=0$.

So, the columns' being linearly independent is nothing but $M(c^1,...,c^n{)}^t=0$'s having only the 0 solution.

By Cramer's rule for any system of linear equations, that is nothing but $detM\ne 0$: if $detM=0$, the rank would be smaller than $n$, and at least 1 of $c^j$ s could be taken arbitrarily.

So, $detM\ne 0$ if and only if the set of the columns is linearly independent.

Step 2:

By Step 1, $det{M}^t\ne 0$ if and only if the set of the columns is linearly independent.

But the set of the columns of $M^t$ is linearly independent if and only if the set of the rows of $M$ is linearly independent, obviously.

On the other hand, $det{M}^t=detM$.

So, $detM\ne 0$ if and only if the set of the rows of $M$ is linearly independent.

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References

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

1306: For Finite-Dimensional Columns or Rows Vectors Space and Linearly Independent Subset, Subset Can Be Shrunk into Number-of-Elements-Dimensional Columns or Rows Vectors Space by Choosing Components

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

description/proof of that for finite-dimensional columns or rows vectors space and linearly independent subset, subset can be shrunk into number-of-elements-dimensional columns or rows vectors space by choosing components

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Topics

About: vectors space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of %field name% vectors space.
• The reader admits the proposition that for any injective linear map between any modules, the image of any linearly independent subset of the domain is linearly independent.
• The reader admits Cramer's rule for any system of linear equations.
• The reader admits the proposition that the determinant of any square matrix over any field is nonzero if and only if the set of the columns or the rows is linearly independent.

---

Target Context

• The reader will have a description and a proof of the proposition that for any finite-dimensional columns or rows vectors space and any linearly independent subset, the subset can be shrunk into a number-of-elements-dimensional columns or rows vectors space by choosing some components.

---

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:
$F$: $\in \{\text{ the fields }\}$
$d^{\prime }$: $\in \mathbb{N}$
$V_c^{\prime }$: $=\{(r^1,...,r^{d^{\prime }}{)}^t|r^j\in F\}$, $\in \{\text{ the }F\text{ vectors spaces }\}$
$V_r^{\prime }$: $=\{(r^1,...,r^{d^{\prime }})|r^j\in F\}$, $\in \{\text{ the }F\text{ vectors spaces }\}$
$S_c^{\prime }$: $=\{({r_1}^1,...,{r_1}^{d^{\prime }}{)}^t,...,({r_n}^1,...,{r_n}^{d^{\prime }}{)}^t\}$, $\in \{\text{ the linearly independent subsets of }{V}_c^{\prime }\}$
$S_r^{\prime }$: $=\{({r_1}^1,...,{r_1}^{d^{\prime }}),...,({r_n}^1,...,{r_n}^{d^{\prime }})\}$, $\in \{\text{ the linearly independent subsets of }{V}_r^{\prime }\}$
$V_c$: $=\{(r^1,...,r^n{)}^t|r^j\in F\}$, $\in \{\text{ the }F\text{ vectors spaces }\}$
$V_r$: $=\{(r^1,...,r^n)|r^j\in F\}$, $\in \{\text{ the }F\text{ vectors spaces }\}$
//

Statements:
$\mathrm{\exists }\{j_1,...,j_n\}\subseteq \{1,...,d^{\prime }\}(S_c:=\{({r_1}^{j_1},...,{r_1}^{j_n}{)}^t,...,({r_n}^{j_1},...,{r_n}^{j_n}{)}^t\}\in \{\text{ the linearly independent subsets of }{V}_c\})$
$\wedge$
$\mathrm{\exists }\{j_1,...,j_n\}\subseteq \{1,...,d^{\prime }\}(S_r:=\{({r_1}^{j_1},...,{r_1}^{j_n}),...,({r_n}^{j_1},...,{r_n}^{j_n})\}\in \{\text{ the linearly independent subsets of }{V}_r\})$
//

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2: Note

It is not that any $\{j_1,...,j_n\}\subseteq \{1,...,d^{\prime }\}$ would do: for example, when $F=\mathbb{R}$, $d^{\prime }=3$, and $S_c^{\prime }=\{(0,1,0{)}^t,(0,0,1{)}^t\}$, $\{1,2\}\subseteq \{1,2,3\}$ does not do because $\{(0,1{)}^t,(0,0{)}^t\}$ is not linearly independent, while $\{2,3\}\subseteq \{1,2,3\}$ does.

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3: Proof

Whole Strategy: Step 1: take $c_1({r_1}^1,...,{r_1}^{d^{\prime }}{)}^t+...+c_n({r_n}^1,...,{r_n}^{d^{\prime }}{)}^t=0$ and see that that equals $\left(\begin{array}{ccc}{r_1}^1& ...& {r_n}^1\\ ...\\ {r_1}^{d^{\prime }}& ...& {r_n}^{d^{\prime }}\end{array}\right)\left(\begin{array}{c}{c}_1\\ ...\\ c_n\end{array}\right)=0$; Step 2: apply Cramer's rule for any system of linear equations to conclude that the rank of the matrix is $n$; Step 3: choose a linearly independent $S_c$; Step 4: conclude for $S_r$.

Step 1:

Let us take $c_1({r_1}^1,...,{r_1}^{d^{\prime }}{)}^t+...+c_n({r_n}^1,...,{r_n}^{d^{\prime }}{)}^t=0$.

That equals $\left(\begin{array}{ccc}{r_1}^1& ...& {r_n}^1\\ ...\\ {r_1}^{d^{\prime }}& ...& {r_n}^{d^{\prime }}\end{array}\right)\left(\begin{array}{c}{c}_1\\ ...\\ c_n\end{array}\right)=0$, obviously.

$S_c$' s being linearly independent equals that equation's having only the $\left(\begin{array}{c}{c}_1\\ ...\\ c_n\end{array}\right)=0$ solution.

Step 2:

By Cramer's rule for any system of linear equations, the rank of the matrix is $n$: if the rank was smaller than $n$, at least 1 of $c_j$ s can be taken arbitrarily, a contradiction.

Step 3:

So, the matrix can be rearranged such that the top-left $n\times n$ submatrix is determinant nonzero.

Then, the $n$ columns of the submatrix is linearly independent, by the proposition that the determinant of any square matrix over any field is nonzero if and only if the set of the columns or the rows is linearly independent.

The $n$ columns is an $S_c$.

Step 4:

As for $S_r$, we can just take the transpose of $S_r^{\prime }$, $S_c^{\prime }\subseteq V_c^{\prime }$, take an $S_c$, and take the transpose of $S_c$, $S_r\subseteq V_r$.

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References

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