Dual Vector Space and Dual Basis - another question - Winitzki Section 1-6

[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 an aspect of the notion or concept of the dual basis \(\displaystyle \{ e^*_1, e^*_2, \ ... \ ... \ , \ e^*_n \} \)

The relevant part of Winitzki's text reads as follows:http://mathhelpboards.com/attachments/linear-abstract-algebra-14/5348-dual-vector-space-dual-basis-another-question-winitzki-section-1-6-a-winitzki-dual-basis-pngIn the above quoted text from Winitzki, we read:

" ... ... Please note that \(\displaystyle e^*_1\) depends on the entire basis \(\displaystyle \{ e^*_j \}\) and not only on \(\displaystyle e^*_1\), as might appear from the notation \(\displaystyle e^*_1\). ... ... "I am puzzled by this statement ... can someone explain how, and indeed why, \(\displaystyle e^*_1\) depends on the entire basis \(\displaystyle \{ e^*_j \}\) and not only on \(\displaystyle e^*_1\) ... a clarification of the nature of the dual basis would be most helpful ...

Hope someone can help ...

Peter===========================================================*** NOTE ***To indicate Winitzki's approach to the dual space (and its basis) and his notation I am providing the text of his introduction to Section 1.6 on the dual or conjugate vector space ... ... as follows ... ...
http://mathhelpboards.com/attachments/linear-abstract-algebra-14/5349-dual-vector-space-dual-basis-another-question-winitzki-section-1-6-a-winitzki-1-section-1-6-part-1-png
http://mathhelpboards.com/attachments/linear-abstract-algebra-14/5350-dual-vector-space-dual-basis-another-question-winitzki-section-1-6-a-winitzki-2-section-1-6-part-2-png

 

[Deveno]

The coordinates of a vector in a basis depend on ALL the basis vectors.

Suppose we have the basis $\{(1,0,0),(0,1,0),(0,0,1)\}$ of $\Bbb R^3$, and we switch to the basis $\{(1,2,3),(0,1,0),(0,0,1)\}$.

Then a given vector which has coordinates $(5,3,1)$ in the standard basis, now has coordinates $(5,-7,-14)$. Even though only ONE basis vector changed, TWO coordinates changed. One can create easily examples where ALL the coordinates change when you change bases.

So if you define a linear functional by choosing the $i$-th coordinate in a given basis, if you change the basis, the same process will yield a DIFFERENT linear functional.

For emphasis: coordinate $n$-tuples are NOT the vectors they represent EXCEPT when one is using the STANDARD basis. A point in $n$-space is just a point in $n$-space. The coordinates are something "we make up" in order to have a handy DESCRIPTION. I may describe someplace as "1 unit" from my house, where by "unit" I mean 2000 feet northwest. If someone else describes it (that same location I refer to) by kilometers east, and kilometers south, the numbers they describe it with will be different than MY "(1,0)" (the "0" refers to, of course, 2000 feet southeast), but we are referring to the same point.

Vector spaces do not come with a basis pre-installed. They have MANY possible bases, and picking one over another is merely a matter of convenience. Again, for emphasis:

The basis $\{(1,0,\dots,0),(0,1,\dots,0),\dots,(0,,\dots,1)\}$ is not "the" basis of $\Bbb R^n$, it just one that allows us to use $n$-tuples INTERCHANGEABLY with the vectors themselves. For some linear transformations, this makes life easier, for others, it does not.

 

[Math Amateur]

The coordinates of a vector in a basis depend on ALL the basis vectors.

Suppose we have the basis $\{(1,0,0),(0,1,0),(0,0,1)\}$ of $\Bbb R^3$, and we switch to the basis $\{(1,2,3),(0,1,0),(0,0,1)\}$.

Then a given vector which has coordinates $(5,3,1)$ in the standard basis, now has coordinates $(5,-7,-14)$. Even though only ONE basis vector changed, TWO coordinates changed. One can create easily examples where ALL the coordinates change when you change bases.

So if you define a linear functional by choosing the $i$-th coordinate in a given basis, if you change the basis, the same process will yield a DIFFERENT linear functional.

For emphasis: coordinate $n$-tuples are NOT the vectors they represent EXCEPT when one is using the STANDARD basis. A point in $n$-space is just a point in $n$-space. The coordinates are something "we make up" in order to have a handy DESCRIPTION. I may describe someplace as "1 unit" from my house, where by "unit" I mean 2000 feet northwest. If someone else describes it (that same location I refer to) by kilometers east, and kilometers south, the numbers they describe it with will be different than MY "(1,0)" (the "0" refers to, of course, 2000 feet southeast), but we are referring to the same point.

Vector spaces do not come with a basis pre-installed. They have MANY possible bases, and picking one over another is merely a matter of convenience. Again, for emphasis:

The basis $\{(1,0,\dots,0),(0,1,\dots,0),\dots,(0,,\dots,1)\}$ is not "the" basis of $\Bbb R^n$, it just one that allows us to use $n$-tuples INTERCHANGEABLY with the vectors themselves. For some linear transformations, this makes life easier, for others, it does not.

Thanks Deveno ... relevant and helpful ...

Still reflecting on what you have said ...

Thank you again ...

Peter