How Does Covector Existence Relate to Tensor Dimensions in Winitzki's Lemma 3?

[Math Amateur]

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

I am currently focused on Section 1.7.3 Dimension of a Tensor Product is the Product of the Dimensions ... ...

I need help in order to get a clear understanding of an aspect of the proof of Lemma 3 in Section 1.7.3 ...

The relevant part of Winitzki's text reads as follows:

In the above quotation from Winitzki we read the following:

" ... ... By the result of Exercise 1 in Sec. 6.3 there exists a covector $f^* \in V^*$ such that

$f^* ( v_j ) = \delta_{ j_1 j }$ for $j = 1, \ ... \ ... \ , \ n$ ... ... "I cannot see how to show that there exists a covector $f^* \in V^*$ such that

$f^* ( v_j ) = \delta_{ j_1 j }$ for $j = 1, \ ... \ ... \ , \ n$ ... ...Can someone help me to show this from first principles ... ?It may be irrelevant to my problem ... but I cannot see the relevance of Exercise 1 in Section 6 which reads as follows:

Exercise 1 refers to Example 2 which reads as follows:

BUT ... since I wish to show the result:

... ... ... "there exists a covector $f^* \in V^*$ such that

$f^* ( v_j ) = \delta_{ j_1 j }$ for $j = 1, \ ... \ ... \ , \ n$ ... ..."... from first principles the above example is irrelevant ... BUT then ... I cannot see its relevance anyway!Hope someone can help ... ...

Peter

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*** NOTE ***

To help readers understand Winitzki's approach and notation for tensors I am providing Winitzki's introduction to Section 1.7 ... ... as follows ... ... :

 

[Samy_A]

In the above quotation from Winitzki we read the following:

" ... ... By the result of Exercise 1 in Sec. 6.3 there exists a covector $f^* \in V^*$ such that

$f^* ( v_j ) = \delta_{ j_1 j }$ for $j = 1, \ ... \ ... \ , \ n$ ... ... "I cannot see how to show that there exists a covector $f^* \in V^*$ such that

$f^* ( v_j ) = \delta_{ j_1 j }$ for $j = 1, \ ... \ ... \ , \ n$ ... ...Can someone help me to show this from first principles ... ?

Read what lavinia explained here:

If one has a basis $e_{i}$ for the vectors space, then a basis for the vector space of covectors - called the dual basis are the linear maps $π_{i}$ defined by $π_{i}(e_{j}) = δ_{ij}$ This is the covector that assigns 1 to the i'th basis vector and zero to all of the others - as mentioned already above. For each choice of basis $e_{i}$ one has a corresponding choice of basis $π_{i}$ for the vector space of covectors.

The covectors $v_{i}$ mentioned above are the same as the covectors $π_{i}$. So the function that picks out the i'th coordinate of a vector with respect to a basis is a covector.

 

[andrewkirk]

In the above quotation from Winitzki we read the following:

" ... ... By the result of Exercise 1 in Sec. 6.3 there exists a covector $f^* \in V^*$ such that

$f^* ( v_j ) = \delta_{ j_1 j }$ for $j = 1, \ ... \ ... \ , \ n$ ... ... "I cannot see how to show that there exists a covector $f^* \in V^*$ such that

$f^* ( v_j ) = \delta_{ j_1 j }$ for $j = 1, \ ... \ ... \ , \ n$ ... ...Can someone help me to show this from first principles ... ?

Firstly, the author's use of notation is unhelpful, because they are using $f^*$ for two very different maps, one from $V$ to $\mathbb{R}$ and one from $V\otimes W$ to $\mathbb{R}$. They should use different symbols for the two different maps.

Let's assume $f^*$ refers only to the first map. We show there exists such a map (which is a covector) simply by defining its operation on a set of basis vectors, and then extending it to cover the whole of the domain $V$ using the linearity rules. To be very clear, let's use a different symbol. Given a basis $B\equiv \{v_1,...,v_n\}$ for $V$, we first define a function $g^{j_1}:B\to\mathbb{R}$ by

$$g^{j_1}(v_k)=\delta_k^{j_1}$$

for $k\in\{1,...,n\}$.

Then we define $f^*$ to be the linear function from $V$ to $\mathbb{R}$ that agrees with $g^{j_1}$ on $B$. It is a basic result of linear algebra (no tensors required) that such a function exists and is unique. If you are not comfortable just accepting that fact, a proof should be easy to find in any decent linear algebra text, or it's easy to prove from scratch. The existence follows from the fact that $\{v_1,...,v_n\}$ spans $V$. To prove uniqueness, assume there are two such extension functions. Then their difference is also a linear function from $V$ to $\mathbb{R}$ and, by looking at its operation on the basis vectors, it's easily seen to be identically zero.

 

[Math Amateur]

Thanks Samy and Andrew ... appreciate the help ...

Still working through the Lemma ... but I must say Andrew, your point on the notation really changed things for me ... I now have a pretty good understanding regarding what is going on in the proof of Lemma 3 ...

Thanks again,

Peter