B Isn't the concept of a basis circular?

[archaic]

If $v$ is an element of a vector space $V$ and for example $\mathcal{B}=\{e_1,e_2,e_3\}$ is a basis of $V$, then, at least, there should be another basis for $V$ in which the vectors of $\mathcal{B}$ can be expressed, but at the same time, the vectors of this other basis must also be expressed using $\mathcal{B}$'s vectors.

Why is there no problem here?

 

[phyzguy]

If $v$ is an element of a vector space $V$ and for example $\mathcal{B}=\{e_1,e_2,e_3\}$ is a basis of $V$, then, at least, there should be another basis for $V$ in which the vectors of $\mathcal{B}$ can be expressed, but at the same time, the vectors of this other basis must also be expressed using $\mathcal{B}$'s vectors.

Why is there no problem here?

Your question is not clear. There are an infinite number of possible bases, and any vector can be expressed in any of the bases. For example, in a 2 dimensional Euclidean space, {ex, ey} is a basis, but I can rotate and find a new basis {ex', ey'} where ex' = √2 ex + √2 ey and ey' = -√2 ex + √2 ey. Any vector, including ex, ey, ex', and ey', can be expressed in either basis. Does this help?

 

[fresh_42]

If $v$ is an element of a vector space $V$ and for example $\mathcal{B}=\{e_1,e_2,e_3\}$ is a basis of $V$, then, at least, there should be another basis for $V$ in which the vectors of $\mathcal{B}$ can be expressed, but at the same time, the vectors of this other basis must also be expressed using $\mathcal{B}$'s vectors.

Why is there no problem here?

Keep it easy. Let's consider $V=\mathbb{R}$. Then every vector different form $0$ is a basis vector and we have bases
$$
\{\,1\,\} \stackrel{\cdot 2}{\longrightarrow} \{\,2\,\} \stackrel{\cdot \frac{3}{2}}{\longrightarrow} \{\,3\,\} \stackrel{\cdot \frac{1}{3}}{\longrightarrow} \{\,1\,\}
$$
Can you rephrase your question with this example?

Of course if we only have $V=\mathbb{F}_2$ we have only one unique basis vector, two in $V=\mathbb{F}_3$ etc., but let's stay with real vector spaces.

 

[archaic]

Keep it easy. Let's consider $V=\mathbb{R}$. Then every vector different form $0$ is a basis vector and we have bases
$$
\{\,1\,\} \stackrel{\cdot 2}{\longrightarrow} \{\,2\,\} \stackrel{\cdot \frac{3}{2}}{\longrightarrow} \{\,3\,\} \stackrel{\cdot \frac{1}{3}}{\longrightarrow} \{\,1\,\}
$$
Can you rephrase your question with this example?

Of course if we only have $V=\mathbb{F}_2$ we have only one unique basis vector, two in $V=\mathbb{F}_3$ etc., but let's stay with real vector spaces.

It seems odd to me that, for example, both $\{1\}$ and $\{2\}$ are basis for each other, there's no building block.

 

[fresh_42]

It seems odd to me that, for example, both $\{1\}$ and $\{2\}$ are basis for each other, there's no building block.

Why that? $b_1=\frac{1}{2}\cdot b_2$ and $b_2=2\cdot b_1$ with two bases $\{\,b_1\,\}$ and $\{\,b_2\,\}$. In this case we have $b_1=1\, , \,b_2=2\,.$

 

[archaic]

Why that? $b_1=\frac{1}{2}\cdot b_2$ and $b_2=2\cdot b_1$ with two bases $\{\,b_1\,\}$ and $\{\,b_2\,\}$. In this case we have $b_1=1\, , \,b_2=2\,.$

And that is not circular? We're basically saying that, this vector space has $\{1\}$ as basis, but that basis is in turn generated by $\{2\}$ which ultimately is also generated by $\{1\}$.
We express vectors of a vector space in terms of a basis' vectors, but those basis' vectors are also vectors and should be expressed with a basis' vectors.

 

[fresh_42]

Of course it is 'circular'. If $\{\,b_i\,\}$ and $\{\,c_i\,\}$ are two basis of an $n-$dimensional vector space, then there is a regular $n\times n$ matrix $A$ such that $Ab_i=c_i$ for all $i$, and as it is regular, we also have $A^{-1}c_i=b_i$ for all $i\,.$

But there are really many regular matrices, so we have many bases.

 

[Stephen Tashi]

And that is not circular?

The mathematical concept of a basis does not embody the idea of "smallest building block" or "atoms". Think of physical 3-D space. There are various ways to specify points in 3D space using 3 basis vectors. There are different ways to define a basis for 3D space.

 

[nuuskur]

If $v$ is an element of a vector space $V$ and for example $\mathcal{B}=\{e_1,e_2,e_3\}$ is a basis of $V$, then, at least, there should be another basis for $V$

This is not required, but there may be. After all, a basis isn't necessarily unique. the initial basis is adequate to express the basis vectors, too: $e_1 = 1\cdot e_1 + 0\cdot e_2 + 0 \cdot e_3$ etc.

If you applied an isomorphism on a basis you would get a basis.