3
$\begingroup$

Some confusion about the basis vectors. These questions came from the definition of a basis at the bottom. I'm wondering:

  1. What the relation is between basis and non-basis vectors. Or put another way, if every single element in the vector space is linearly dependent on the linearly independent basis vectors (they are independent among themselves, but dependent amongst the other vectors). This is based on my understanding of the definition of span and the idea that every vector in the space is a linear combination of the basis vectors (except the basis vectors). I'm visualizing the span as: the basis vectors are all linearly independent, but then they "connect" to every other vector outside of the basis via linear combination. In that sense they reach every other vector, and so span the vector space. Wondering if that is correct.
  2. If the number of basis vectors is the dimension. Just want to make sure I'm understanding.
  3. What the relation between a basis and a generating set is.
  4. A brief intuition on how to think of basis vectors.

A set of elements in a vector space is called a basis if they are:

  1. Linearly independent.
  2. And every vector in the vector space is a linear combination of the set (the basis vectors?).

In another perspective, a basis is a linearly independent spanning set.

Formally, a basis $B$ of a vector space $V$ over a field $F$ is a linearly independent subset of $V$ that spans $V$.

A span of a set of vectors in $V$ is the intersection of all subspaces containing that set.

In the definition of free module they compare a basis to a generating set.

$\endgroup$
  • $\begingroup$ It seems the problem might be that there are many choices of a basis for a given vector space. So your definition of dimension in 2 is referring to number of basis vectors in a particular basis, not every vector that is ostensibly a ‘basis vector’ $\endgroup$ – Prince M May 15 '18 at 8:05
  • $\begingroup$ You can think of basis vectors as a minimal spanning set. It is the fewest number of vectors needed to span the entire vector space. $\endgroup$ – Prince M May 15 '18 at 8:07
  • $\begingroup$ In the case of R^2 you can draw pictures to get intuition and see how any two linearly independent vectors will let you reach any other point in the space, but In higher dimensions or abstract vector spaces there may not be geometric intuition $\endgroup$ – Prince M May 15 '18 at 8:09
5
$\begingroup$

Let's talk about vectors of real numbers with two elements in them. The set of these vectors is named $\mathbb{R}^2$. Consider $$\begin{bmatrix}1\\ 0\end{bmatrix} \text{and} \begin{bmatrix}0\\ 1\end{bmatrix}.$$

Can any other two element vector be written as a combination of these two vectors? Yes, it can! Consider any vector $\begin{bmatrix}a\\ b\end{bmatrix}$ for any values $a$ and $b$. $\begin{bmatrix}a\\ b\end{bmatrix}=a\begin{bmatrix}1\\ 0\end{bmatrix} + b\begin{bmatrix}0\\ 1\end{bmatrix}.$ Since any two element vector can be written as a linear combination of these two vectors, and we need both of these vectors, then these vectors form a basis.

Formally, we say that $\left\{\begin{bmatrix}1\\ 0\end{bmatrix},\begin{bmatrix}0\\ 1\end{bmatrix}\right\}$ is a basis of $\mathbb{R}^2$. We could also say that $\left\{\begin{bmatrix}1\\ 0\end{bmatrix},\begin{bmatrix}0\\ 1\end{bmatrix}\right\}$ generates $\mathbb{R}^2$.

What about the vectors $$\begin{bmatrix}1\\ 0\end{bmatrix}, \begin{bmatrix}0\\ 1\end{bmatrix} \text{and} \begin{bmatrix}1\\ 1\end{bmatrix}?$$ We could also write any vector $\begin{bmatrix}a\\ b\end{bmatrix}$ as a linear combination of these three vectors. However, we don't need all three of them. We could just as easily work with just the first two vectors. So the set of these vectors is not a basis of $\mathbb{R}^2$. However, it still generates $\mathbb{R}^2$ because $\text{span}\left(\left\{\begin{bmatrix}1\\ 0\end{bmatrix},\begin{bmatrix}0\\ 1\end{bmatrix},\begin{bmatrix}1\\ 1\end{bmatrix}\right\}\right)=\mathbb{R}^2$. Therefore, the dimension of $\mathbb{R}^2$ is two, because any basis set of $\mathbb{R}^2$ has two vectors in it.

Can we have a basis other than $\left\{\begin{bmatrix}1\\ 0\end{bmatrix},\begin{bmatrix}0\\ 1\end{bmatrix}\right\}$? Yes, we can. Consider $$\begin{bmatrix}1\\ 1\end{bmatrix} \text{and} \begin{bmatrix}0\\ 1\end{bmatrix}.$$ Any vector $\begin{bmatrix}a\\ b\end{bmatrix}$ can be written as a linear combination of these two vectors: $$ a\begin{bmatrix}1\\ 1\end{bmatrix} + (b-a)\begin{bmatrix}0\\ 1\end{bmatrix}=\begin{bmatrix}a\\ b\end{bmatrix}.$$ So $\left\{\begin{bmatrix}1\\ 1\end{bmatrix},\begin{bmatrix}0\\ 1\end{bmatrix}\right\}$ is also a basis of $\mathbb{R}^2$.

Imagine trying to send a vector space through a communication channel. If we had to send every vector then this would be an impossible thing to do because we would need an infinite number of vectors. What could we do instead? If we had a basis for our vector space, we could just send the basis and tell the person on the other end that the vectors we sent are a basis for our vector space. And that's the intuition. A basis is a small set of vectors that completely represents our vector space. In fact, there's no smaller set of vectors that completely represents our vector space.

$\endgroup$
2
$\begingroup$
  1. There is nothing that distinguishes basis vectors from non-basis vectors. Any vector (except the zero vector) can be part of a basis. There are many bases for a given vector space.

  2. The dimension of a vector space is defined as the cardinality of a basis.

  3. A basis is a generating set that happens to be linearly independent. If you add additional vectors to a basis, it will still be a generating set, but it will no longer be linearly independent.

  4. Don't think of basis vectors. Rather than thinking of the vectors in the basis as special, think of the basis itself (a collection of vectors) as being special.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.