MHB Vector Spaces - The Exchange Theorem - Cooperstein Theorem 1.16

[Math Amateur]

I am reading Bruce Cooperstein's book: Advanced Linear Algebra ... ...

I am focused on Section 1.6 Bases and Finite-Dimensional Vector Spaces ...

I need help with the proof of Theorem 1.16 ...

Theorem 1.16 and its proof reads as follows:
View attachment 5134
Question 1

In the second paragraph of above proof we read the following:

"... ... Since $$y_1 \neq 0$$ it follows ... ... Can someone please explain exactly why $$y_1 \neq 0$$?Question 2

In the second paragraph of above proof we read the following:

"... ... By reordering the vectors of $$X$$, we can assume that $$x_n$$ is a linear combination of $$Z_1 = ( y_1, x_1, \ ... \ ... \ , x_{n-1}$$. ... ... "

Can someone explain (in some detail) exactly why $$x_n$$ is a linear combination of $$Z_1 = ( y_1, x_1, \ ... \ ... \ , x_{n-1}) $$?
Hope someone can help ... ...

Help will be much appreciated ... ...

Peter
NOTE: The text from Cooperstein displayed above mentions Theorem 1.14 ... so I am providing the statement of Theorem 1.14 as follows:View attachment 5135

 

[Fallen Angel]

Hi Peter,

Can someone please explain exactly why $$y_1 \neq 0$$?

In the proof, we are assuming that $Y$ is linearly independent, so it can't contain null vectors.

Can someone explain (in some detail) exactly why $$x_n$$ is a linear combination of $$Z_1 = ( y_1, x_1, \ ... \ ... \ , x_{n-1}$$?

Once you have noted that $y_{1}\in Span(X)$ you hav that $y_{1}=\displaystyle\sum_{i=0}^{m}c_{i}x_{i}$ for some $c_{i}\in K$ (being $K$ the field under your vector space $V$).
Now, since $y_{1}\neq 0$ there is at least one non zero coefficient in the above sum, say $c_{j}$.
Can you then imagine how $x_{j}$ is a linear combimation of $x_{i}$, with $i\neq j$ and $y_{1}$?
If you can do that, the rest is just taking a different order in the indices.

 

[ThePerfectHacker]

Can someone please explain exactly why $$y_1 \neq 0$$?

By assumption you start with a lin. ind. set, therefore no vector in it can be zero. (Or show that if you have a list of vectors, and a zero vector is among them, then that list is automatically lin. dep.)

 

[Deveno]

It is crucial, when talking about bases, to thoroughly wrap one's mind around the concept of linear independence, and linear dependence.

A finite set $S = \{v_1,v_2,\dots,v_k\} \subset V$ of vectors in a vector space $V$ over a field $F$ is said to be linear dependent if there exist scalars $c_1,c_2,\dots,c_k \in F$ not all zero such that:

$c_1v_1 + c_2v_2 + \cdots + c_kv_k = 0_V$

In particular, if $c_j \neq 0_F$, then:

$v_j = -\dfrac{c_1}{c_j}v_1 - \cdots -\dfrac{c_{j-1}}{c_j}v_{j-1} - \dfrac{c_{j+1}}{c_j}v_{j+1}-\cdots-\dfrac{c_k}{c_j}v_k$

that is, $v_j \in \text{span}(S - \{v_j\})$, so we have some "redundancy" in $S$ as a spanning set of $\text{span}(S)$.

As an easy consequence, we can prove, straight from the definition:

for any finite set $S \subset V$ such that $0_V \in S$, $S$ is a linearly dependent set.

Proof: re-label the set $S$ like so: $\{v_1,v_2,\dots,v_{k-1},0_V\}$. Choose $\{c_1,c_2,\dots,c_{k-1},c_k\}$ in the following way:

$c_1 = c_2 = \cdots = c_{k-1} = 0_F$, and $c_k = 1_F$.

Since $F$ is a field, $0_F \neq 1_F$, and our set of $c$'s is not all zeros (it has one non-zero element).

Then $c_1v_1 + \cdots + c_{k-1}v_{k-1} + c_k 0_V = 0_Fv_1 + \cdots + 0_Fv_{k-1} + 1_F0_V$

$= 0_V +\cdots + 0_V + 1_F0_V = 1_F0_V = 0_V$.

Linear independence is the negation of the property of linear dependence. Since linear dependence asserts the existence of a certain set of non-zero scalars that produce a 0 linear combination, linear independence asserts that for all such non-zero sets, they never produce a 0 linear combination. We can restate this as:

A finite set $S = \{v_1,\dots,v_k\} \subset V$ is linearly independent if:

$c_1v_1 + \cdots + c_kv_k = 0_V \implies c_1 = \cdots = c_k = 0_F$.

If a (finite) set $S$ is linearly independent, *all* the elements of $S$ are necessary to generate $\text{span}(S)$. For if $v_j \in S$, there is no way to realize $v_j$ as a linear combination of the other elements of $S$ (for if so, if we could write:

$v_j = d_1v_1 + \cdots + d_{j-1}v_{j-1} + d_{j+1}v_{j+1} + \cdots + d_kv_k$

then we would have:

$d_1v_1 + \cdots + d_{j-1}v_{j-1} - 1_Fv_j + d_{j+1}v_{j+1} + \cdots + d_kv_k = 0_V$, and the set:

$\{d_1,\dots,-1_F,d_{j+1},\dots,d_k\}$ is not all zero elements of $F$ ($-1_F$ cannot be zero, do you see why?). This contradicts the linear independence of $S$).

These ideas can be extended to infinite sets, as well, if we only allow *finite* linear combinations. But there are certain subtleties involved which are better left alone, for now (for finite-dimensional vector spaces, you will not need to examine infinite generating sets).

 

[Math Amateur]

By assumption you start with a lin. ind. set, therefore no vector in it can be zero. (Or show that if you have a list of vectors, and a zero vector is among them, then that list is automatically lin. dep.)

Thanks Fallen Angel and ThePerfectHacker ... appreciate the help ...

Peter

- - - Updated - - -

It is crucial, when talking about bases, to thoroughly wrap one's mind around the concept of linear independence, and linear dependence.

A finite set $S = \{v_1,v_2,\dots,v_k\} \subset V$ of vectors in a vector space $V$ over a field $F$ is said to be linear dependent if there exist scalars $c_1,c_2,\dots,c_k \in F$ not all zero such that:

$c_1v_1 + c_2v_2 + \cdots + c_kv_k = 0_V$

In particular, if $c_j \neq 0_F$, then:

$v_j = -\dfrac{c_1}{c_j}v_1 - \cdots -\dfrac{c_{j-1}}{c_j}v_{j-1} - \dfrac{c_{j+1}}{c_j}v_{j+1}-\cdots-\dfrac{c_k}{c_j}v_k$

that is, $v_j \in \text{span}(S - \{v_j\})$, so we have some "redundancy" in $S$ as a spanning set of $\text{span}(S)$.

As an easy consequence, we can prove, straight from the definition:

for any finite set $S \subset V$ such that $0_V \in S$, $S$ is a linearly dependent set.

Proof: re-label the set $S$ like so: $\{v_1,v_2,\dots,v_{k-1},0_V\}$. Choose $\{c_1,c_2,\dots,c_{k-1},c_k\}$ in the following way:

$c_1 = c_2 = \cdots = c_{k-1} = 0_F$, and $c_k = 1_F$.

Since $F$ is a field, $0_F \neq 1_F$, and our set of $c$'s is not all zeros (it has one non-zero element).

Then $c_1v_1 + \cdots + c_{k-1}v_{k-1} + c_k 0_V = 0_Fv_1 + \cdots + 0_Fv_{k-1} + 1_F0_V$

$= 0_V +\cdots + 0_V + 1_F0_V = 1_F0_V = 0_V$.

Linear independence is the negation of the property of linear dependence. Since linear dependence asserts the existence of a certain set of non-zero scalars that produce a 0 linear combination, linear independence asserts that for all such non-zero sets, they never produce a 0 linear combination. We can restate this as:

A finite set $S = \{v_1,\dots,v_k\} \subset V$ is linearly independent if:

$c_1v_1 + \cdots + c_kv_k = 0_V \implies c_1 = \cdots = c_k = 0_F$.

If a (finite) set $S$ is linearly independent, *all* the elements of $S$ are necessary to generate $\text{span}(S)$. For if $v_j \in S$, there is no way to realize $v_j$ as a linear combination of the other elements of $S$ (for if so, if we could write:

$v_j = d_1v_1 + \cdots + d_{j-1}v_{j-1} + d_{j+1}v_{j+1} + \cdots + d_kv_k$

then we would have:

$d_1v_1 + \cdots + d_{j-1}v_{j-1} - 1_Fv_j + d_{j+1}v_{j+1} + \cdots + d_kv_k = 0_V$, and the set:

$\{d_1,\dots,-1_F,d_{j+1},\dots,d_k\}$ is not all zero elements of $F$ ($-1_F$ cannot be zero, do you see why?). This contradicts the linear independence of $S$).

These ideas can be extended to infinite sets, as well, if we only allow *finite* linear combinations. But there are certain subtleties involved which are better left alone, for now (for finite-dimensional vector spaces, you will not need to examine infinite generating sets).

Thank you, Deveno, for a particularly helpful post ... really appreciate your help ...

Peter

 

[Math Amateur]

Thanks Fallen Angel and ThePerfectHacker ... appreciate the help ...

Peter

- - - Updated - - -Thank you, Deveno, for a particularly helpful post ... really appreciate your help ...

Peter

There is another point that I am not quite clear on regarding Cooperstein's proof ... so for convenience of MHB readers I will provide the text of the Exchange Theorem and its proof again ...https://www.physicsforums.com/attachments/5138In the above text, we read the following:

"... ... Since we are assuming that $$x_n \in \text{Span} (Z_1)$$, it follows that $$\text{Span} (Z_1) = \text{Span} (X) = V$$ ... ... Can someone please explain exactly why $$x_n \in \text{Span} (Z_1)$$ implies that $$\text{Span} (Z_1) = \text{Span} (X) = V$$?

Help will be appreciated ... ...

Peter

 

[Deveno]

There is another point that I am not quite clear on regarding Cooperstein's proof ... so for convenience of MHB readers I will provide the text of the Exchange Theorem and its proof again ...In the above text, we read the following:

"... ... Since we are assuming that $$x_n \in \text{Span} (Z_1)$$, it follows that $$\text{Span} (Z_1) = \text{Span} (X) = V$$ ... ... Can someone please explain exactly why $$x_n \in \text{Span} (Z_1)$$ implies that $$\text{Span} (Z_1) = \text{Span} (X) = V$$?

Help will be appreciated ... ...

Peter

If $\{y_1,x_1,\dots,x_n\}$ is linearly dependent, there is some NOT ALL ZERO $c_j$ for $j = 1,2,\dots,n,n+1$ such that:

$c_1x_1 + c_2x + \cdots + c_nx_n + c_{n+1}y_1 = 0$

Now, if $c_j = 0$ for all $j = 1,2,\dots,n$ (only the first $n$), we are left with:

$c_{n+1}y_1 = 0$, and since our $c_j$ were assumed not ALL zero, we would have $c_{n+1} \neq 0$, forcing $y_1 = 0$.

But $0$ cannot be part of any linearly independent set (see my previous post), so this possibility violates our assumption that $Y = \{y_1,\dots,y_{n+1}\}$ is linearly independent.

So for $Y$ to be linearly independent, we have to have one of the "earlier" $c_j$ be non-zero. It doesn't really matter which one, for the sake of ease in writing the subscripts, we assume without loss of generality it is $c_n$ (if it isn't actuallty $c_n$, we can just re-arrange the $x_i$).

But then this gives:

$c_nx_n = -c_1x_1 - \cdots - c_{n-1}x_{n-1} - c_{n+1}y_1$

and since $c_n \neq 0$, we can divide both sides by it, to explicitly get $x_n$ as a linear combination of $\{y_1,x_1,\dots,x_{n-1}\}$.

Thus any element of $\text{span}(X)$, that is:

$a_1x_1 + \cdots + a_nx_n$ is equal to:

$a_1x_1 + \cdots + a_{n-1}x_{n-1} + a_n\left( -\dfrac{c_1}{c_n}x_1 - \cdots - \dfrac{c_{n-1}}{c_n}x_{n-1} -\dfrac{ c_{n+1}}{c_n}y_1\right)$

$= \left(a_1 - \dfrac{a_nc_1}{c_n}\right)x_1 + \cdots + \left(a_{n-1} - \dfrac{a_nc_{n-1}}{c_n}\right)x_{n-1} - \dfrac{a_nc_{n+1}}{c_n}y_1$,

which is an element of $\text{span}(Z_1)$, so $\text{span}(X) \subseteq \text{span}(Z_1)$.

The other inclusion is obvious, since $\text{span}(X) = V$, and any linear combination of elements of $V$ is contained in $V$, by the closure axioms of a vector space.

 

[Math Amateur]

If $\{y_1,x_1,\dots,x_n\}$ is linearly dependent, there is some NOT ALL ZERO $c_j$ for $j = 1,2,\dots,n,n+1$ such that:

$c_1x_1 + c_2x + \cdots + c_nx_n + c_{n+1}y_1 = 0$

Now, if $c_j = 0$ for all $j = 1,2,\dots,n$ (only the first $n$), we are left with:

$c_{n+1}y_1 = 0$, and since our $c_j$ were assumed not ALL zero, we would have $c_{n+1} \neq 0$, forcing $y_1 = 0$.

But $0$ cannot be part of any linearly independent set (see my previous post), so this possibility violates our assumption that $Y = \{y_1,\dots,y_{n+1}\}$ is linearly independent.

So for $Y$ to be linearly independent, we have to have one of the "earlier" $c_j$ be non-zero. It doesn't really matter which one, for the sake of ease in writing the subscripts, we assume without loss of generality it is $c_n$ (if it isn't actuallty $c_n$, we can just re-arrange the $x_i$).

But then this gives:

$c_nx_n = -c_1x_1 - \cdots - c_{n-1}x_{n-1} - c_{n+1}y_1$

and since $c_n \neq 0$, we can divide both sides by it, to explicitly get $x_n$ as a linear combination of $\{y_1,x_1,\dots,x_{n-1}\}$.

Thus any element of $\text{span}(X)$, that is:

$a_1x_1 + \cdots + a_nx_n$ is equal to:

$a_1x_1 + \cdots + a_{n-1}x_{n-1} + a_n\left( -\dfrac{c_1}{c_n}x_1 - \cdots - \dfrac{c_{n-1}}{c_n}x_{n-1} -\dfrac{ c_{n+1}}{c_n}y_1\right)$

$= \left(a_1 - \dfrac{a_nc_1}{c_n}\right)x_1 + \cdots + \left(a_{n-1} - \dfrac{a_nc_{n-1}}{c_n}\right)x_{n-1} - \dfrac{a_nc_{n+1}}{c_n}y_1$,

which is an element of $\text{span}(Z_1)$, so $\text{span}(X) \subseteq \text{span}(Z_1)$.

The other inclusion is obvious, since $\text{span}(X) = V$, and any linear combination of elements of $V$ is contained in $V$, by the closure axioms of a vector space.

Thanks for for the help Deveno ... much appreciated ...

... working through the the detail of your post now ...

Thanks again,

Peter