# Linear Space Shilov: Subspace Basis & Contradictions

• Locrianz
In summary: It is a bit hard to read your description, some details are missing, maybe I assumed some things that weren't. I don't see how the first statement is true with this information you gave, just pick a vector from L in K so there is something missing. Or I am just dumb and missing something you wrote.In summary, Shilov says that a subspace, say L, must have dimension no greater than 'n' for an 'n' dimensional linear(vector) space K. He goes on to say that if a basis is chosen in the subspace L, say f1,f2,...,fr, then additional vectors can be chosen from the linear space K, say f(r+1),
Locrianz
Hi guys i just had a quick question re shilov linear algebra page 44 here he shows that for an 'n' dimensional linear(vector) space K, a subspace, say L, must have dimention no greater than 'n'.
He then goed onto say that if a basis is chosen in the subspace L, say f1,f2,...,fr, then additional vectors can be chosen from the linear space K, say f(r+1),...,fn such that f1,f2,...,fr,...,fn form a basis for K. Which makes perfect sense. But here is the part that throws me:
Suppose we have the relation:
c1f1+c2f2+...+crfr+c(r+1)f(r+1)=0
Where the c terms are constants and f(r+1) is an additional vector from K added to the basis of L.
He says if c(r+1) does not equal zero then then L is K. Which is true. But then he says if c(r+1)=0 then the vectors f1,f2,...,fr are linearly dependent, which would be a contradiction. How can this be true? If this is the basis for L and we set it equal to zero then we have all the constants equal to zero, hence they will be linearly independent. I checked the errata and there was no mention of this.
Any help will be appreciated, thanks guys.

Locrianz said:
He says if c(r+1) does not equal zero then then L is K. Which is true. But then he says if c(r+1)=0 then the vectors f1,f2,...,fr are linearly dependent, which would be a contradiction. How can this be true? If this is the basis for L and we set it equal to zero then we have all the constants equal to zero, hence they will be linearly independent. I checked the errata and there was no mention of this.
Any help will be appreciated, thanks guys.
Read this line again, f(r+1) is not included in the basis for L:
Locrianz said:
Where the c terms are constants and f(r+1) is an additional vector from K added to the basis of L.

Yes but he say if you set the f(r+1)'s constant equal to zero then the remaining vectors would be linearly dependent? If you set this constant(c(r+1)) equal to zero wouldn't the f(r+1) disappear as its coefficient is now zero and you are just left with the basis for L set equal to zero? Maybe I am missing something but that's the way i see it...

Locrianz said:
Yes but he say if you set the f(r+1)'s constant equal to zero then the remaining vectors would be linearly dependent? If you set this constant(c(r+1)) equal to zero wouldn't the f(r+1) disappear as its coefficient is now zero and you are just left with the basis for L set equal to zero? Maybe I am missing something but that's the way i see it...
It is a bit hard to read your description, some details are missing, maybe I assumed some things that weren't. I don't see how the first statement is true with this information you gave, just pick a vector from L in K so there is something missing. Or I am just dumb and missing something you wrote.

I think it may just be an error, the book was translated from russian so maybe that's an issue, he usually doesn't give much detail, but here he is trying to construct the basis for a vector space K from the basis of the subspace L by adding vectors that are not spanned by the basis of L. And he uses the reasoning above as a step in his proof, but no matter how i think of it it doesn't make sense.

Is he trying to show that $\{f_1,...,f_{n+1}\}$ is linear independent?

In that case he supposes that that

$$c_1f_1+...+c_nf_n+c_{n+1}f_{n+1}=0$$

and that not all ck are 0. The aim is to get a contradiction.

and suppose that cn+1=0, then

$$c_1f_1+...+c_nf_n=0$$

Since the $\{f_1,...,f_n\}$ forms a basis, we see that $c_1=...=c_n=0$, so all the ck are 0, which is a contradiction.

Ah yes that would make sense! Your a genius! He didnt mention explicitly that we should assume that not all the coefficients are zero, but its implict in there since he draws a contradiction, and your explanation is the only way this makes sense! Thanks alot! I can't believe you figured that out with so little information! Cheers.

## 1. What is a linear space?

A linear space, also known as a vector space, is a mathematical concept that represents a collection of objects, called vectors, that can be added together and multiplied by a scalar to create new vectors. It follows a set of axioms and is often used in fields such as physics, engineering, and economics.

## 2. What is a subspace?

A subspace is a subset of a linear space that also follows the axioms of a linear space. In other words, it is a smaller vector space that exists within a larger vector space. Subspaces are useful for simplifying complex problems and finding solutions in a more efficient manner.

## 3. What is a basis?

In linear algebra, a basis is a set of linearly independent vectors that can be used to represent any vector in a linear space. This means that any vector in the linear space can be written as a linear combination of the basis vectors. A basis is important because it allows us to uniquely describe and manipulate vectors in a linear space.

## 4. What are contradictions in linear spaces?

Contradictions, also known as inconsistencies, occur in linear spaces when a set of vectors fails to follow the axioms of a linear space. This means that the set of vectors cannot be considered a linear space and therefore cannot be used in calculations or solutions. Contradictions can arise when there are errors in calculations or when a set of vectors is not properly defined.

## 5. What is the Shilov boundary?

The Shilov boundary, named after Russian mathematician German Shilov, is a set of boundary points that separates a linear space into two disjoint sets. These sets are known as the interior and exterior of the boundary. The Shilov boundary is useful for studying the properties of linear spaces and can aid in solving problems related to linear spaces.

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