MHB How Are Coefficients of a Vector Linear Functions of the Vector?

[Math Amateur]

I am reading Segei Winitzki's book: Linear Algebra via Exterior Products ...

I am currently focused on Section 1.6: Dual (conjugate) vector space ... ...

I need help in order to get a clear understanding of the notion or concept of coefficients of a vector $$v$$ as linear functions (covectors) of the vector $$v$$ ...

The relevant part of Winitzki's text reads as follows:
View attachment 5344In the above text we read:" ... ... So the coefficients $$v_k, \ 1 \leq k \leq n$$, are linear functions of the vector $$v$$ ; therefore they are covectors ... ... "Now, how and in what way exactly are the coefficients $$v_k$$ a function of the vector $$v$$ ... ... ? To indicate my confusion ... if the coefficient $$v_k$$ is a linear function of the vector $$v$$ then $$v_k(v)$$ must be equal to something ... but what? ... indeed what does $$v_k(v)$$ mean? ... further, what, if anything, would v_k(w) mean where w is any other vector? ... and further yet, how do we formally and rigorously prove that $$v_k$$ is linear? ... what would the formal proof look like?... ...

Hope someone can help ...

Peter

===========================================================*** NOTE ***To indicate Winitzki's approach to the dual space and his notation I am providing the text of his introduction to Section 1.6 on the dual or conjugate vector space ... ... as follows ... ... View attachment 5345
View attachment 5346

 

[Deveno]

Re: Coefficients of a vector regarded as a fiunction of a vector ... clarification needed ...

The way I am used to seeing this "co-vector" defined is like so:

Suppose $v = \sum\limits_j v_je_j$, where $\{e_j\}$ is a basis (perhaps the standard basis, perhaps not). We define:

$\pi_i(v) = v_i$

(Note we have as many $\pi$-functions, as we have coordinates).

Thus $\pi_i: V \to F$, since $v$ is a vector, and $v_i$ is a scalar.

For EACH $i$, we have $\pi_i$ is linear: suppose $u = \sum\limits_j u_je_j$ and $v = \sum\limits_j v_je_j$.

Then $u+v = \sum\limits_j (u_j+v_j)e_j$ because for any vector $v$, and any two scalars $a,b$ we have:

$(a+b)v = av + bv$ (this is one of the axioms that define a vector space).

In particular, this is true for each of the basis vectors $e_j$.

So $\pi_i(u+v) = u_i+v_i = \pi_i(u) + \pi_i(v)$.

We also have, for any $a \in F$, that $a\left(\sum\limits_j v_je_j\right) = \sum\limits_j (av_j)e_j$, since for any two vectors $u,v$, and any scalar $a$ we have:

$a(u+v) = au + av$ (this is another vector space axiom), so we can distribute the $a$ over the linear combination of basis vectors.

So $\pi_i(av) = av_i = a(\pi_i(v))$, and we have a linear function.

Note that Winitzki is just naming the function by its image, something that is often done with functions (we often talk about "the function $x^2$" when what we really MEAN is "the squaring function"). What he really means is the function:

$v \mapsto v_i$ (function that returns the $i$-th coordinate of $v$ in some basis).

It is also important to note here that the function(s) we have defined here *depend on a choice of basis*, because the CO-ORDINATES of a vector depend on the basis used.

Put another way, the choice of a basis $\{e_j\}$ creates an isomorphism $\phi:V \to F^{\dim(V)}$, the $\dim(V)$-fold direct sum of $F$. Explicitly, this isomorphism is:

$\phi(v_1e_1 + \cdots + v_ne_n) = (v_1,\dots,v_n)$.

GIVEN the basis $\{e_j\}$ we can specify a *dual basis* $\{\epsilon_i\}$ by:

$\epsilon_i(e_j) = \delta_{ij}$ (this is 1 if $i = j$, and 0 otherwise). It is not hard to see that $\epsilon_i = \pi_i$, for each $i$, by examining $\pi_i(e_j)$ and extending by linearity.

I have tried to illustrate what is happening here by using "different letters" for the $i$-th coordinate function, and the $i$-th coordinate.

 

[Math Amateur]

Re: Coefficients of a vector regarded as a fiunction of a vector ... clarification needed ...

The way I am used to seeing this "co-vector" defined is like so:

Suppose $v = \sum\limits_j v_je_j$, where $\{e_j\}$ is a basis (perhaps the standard basis, perhaps not). We define:

$\pi_i(v) = v_i$

(Note we as many $\pi$-functions, as we have coordinates).

Thus $\pi_i: V \to F$, since $v$ is a vector, and $v_i$ is a scalar.

For EACH $i$, we have $\pi_i$ is linear: suppose $u = \sum\limits_j u_je_j$ and $v = \sum\limits_j v_je_j$.

Then $u+v = \sum\limits_j (u_j+v_j)e_j$ because for any vector $v$, and any two scalars $a,b$ we have:

$(a+b)v = av + bv$ (this is one of the axioms that define a vector space).

In particular, this is true for each of the basis vectors $e_j$.

So $\pi_i(u+v) = u_i+v_i = \pi_i(u) + \pi_i(v)$.

We also have, for any $a \in F$, that $a\left(\sum\limits_j v_je_j\right) = \sum\limits_j (av_j)e_j$, since for any two vectors $u,v$, and any scalar $a$ we have:

$a(u+v) = au + av$ (this is another vector space axiom), so we can distribute the $a$ over the linear combination of basis vectors.

So $\pi_i(av) = av_i = a(\pi_i(v))$, and we have a linear function.

Note that Winitzki is just naming the function by its image, something that is often done with functions (we often talk about "the function $x^2$" when what we really MEAN is "the squaring function"). What he really means is the function:

$v \mapsto v_i$ (function that returns the $i$-th coordinate of $v$ in some basis).

It is also important to note here that the function(s) we have defined here *depend on a choice of basis*, because the CO-ORDINATES of a vector depend on the basis used.

Put another way, the choice of a basis $\{e_j\}$ creates an isomorphism $\phi:V \to F^{\dim(V)}$, the $\dim(V)$-fold direct sum of $F$. Explicitly, this isomorphism is:

$\phi(v_1e_1 + \cdots + v_ne_n) = (v_1,\dots,v_n)$.

GIVEN the basis $\{e_j\}$ we can specify a *dual basis* $\{\epsilon_i\}$ by:

$\epsilon_i(e_j) = \delta_{ij}$ (this is 1 if $i = j$, and 0 otherwise). It is not hard to see that $\epsilon_i = \pi_i$, for each $i$, by examining $\pi_i(e_j)$ and extending by linearity.

I have tried to illustrate what is happening here by using "different letters" for the $i$-th coordinate function, and the $i$-th coordinate.

Well! ... thanks so much Deveno ... that was extremely helpful ...

Indeed while I can see straight away that $$\pi_i$$ is a function ... I was having real trouble with Winitzki asserting that $$v_i$$ was a linear function of $$v$$ (or at least could be understood as a linear function of $$v$$) ...

But then ... the key for me was your explanation ...

" ... ... Note that Winitzki is just naming the function by its image, something that is often done with functions (we often talk about "the function $x^2$" when what we really MEAN is "the squaring function"). What he really means is the function:

$v \mapsto v_i$ (function that returns the $i$-th coordinate of $v$ in some basis). ...

... ... ... "

I still feel a bit uncomfortable with this understanding ... but at least I understand what Winitzki means ... I note that you say this is often done ... I think another example is when we identify the Jacobian matrix with the total derivative or differential in the analysis of vector functions ...

Thanks again,

Peter