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I have a brief understanding of bases. But I don't know if it is right or not. So, I just need someone to correct me if it's not.

When we look for the basis of the image of a matrix, we simply remove all the redundant vectors from the matrix, and keep the linearly independent column vectors. When we look for the basis of the kernel of a matrix, we remove all the redundant column vectors from the kernel, and keep the linearly independent column vectors.

Therefore, a basis is just a combination of all the linearly independent vectors.

By the way, is basis just the plural form of base?

Let me know if I am right.

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Yes, essentially a basis is a set ( not a ''combination'', that is a word without a well defined meaning) of linearly independent vectors that span a vector space.

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What is a basis?

Informally we say

A basis is a set of vectors that generates all elements of the vector space and the vectors in the set are linearly independent.

This is what we mean when creating the definition of a base. It is useful understand an relation between all vectors of the space. They all will have something in common: they can be write as a linear combination of some set of vectors that lies in the space. The set of vectors are called the base of the vector space.

How to make this notion formal?

For that we use the theory of linear algebra. We define what is a vector and what we mean by a vector been generated by other vectors. We say that if a vector is some linear combination of other vectors - with respect to elements of some field (a vector space must have a field in the definition, usually this field is $\mathbb{R}$ or $\mathbb{C}$) - then this vector is generated. In some sense then we find first the set off vectors that generates all vectors in space (can be an infinite of finite set).

Then the theory of linearly independence plays the key role: Two vectors can generate and be generated by other vectors. So we talk about linearly independence when we want that the set that generates the space become

The smallest set of vectors that generates the space

So if I have a set of vectors that generates the space and one - or more - of these vectors is generated by other vectors, then I take this vector out of the set. And in some sense, given that I have alredy a base of my space, if I take out some vector of my base then I cannot generate all vector space anymore! For example we have $\mathbb{R}^2$ and the basis vectors $(0,1)$ and $(1,0)$; we cannot generate $(0,1)$ by a linear combination of $(1,0)$. But of course we can generate all vectors $(a,0)$ for $a \in \mathbb{R}$ using the vector $(1,0)$.

But this is not an unique notion !

It is not! An vector space can have multiple different basis. For example we have for $\mathbb{R}^2$ we have that $\{(1,0),(0,1)\}$ is a basis and we also get that $\{(3,0),(0,5)\}$ is also a basis. But the important notion is that we can create all vectors (poins) in the space $\mathbb{R}^2$ using these sets. So in some sense the basis tells us important things about the space: tells a relation between all vectors; tells how to create an vector; say how we can introduce more profound things about the space such as linear transformations between vector spaces that are different.

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    $\begingroup$ Since it is not unique you shouldn't say "the basis" to begin with. Also, your informal treatment will tend to blur the line between a spanning set and a basis. $\endgroup$ – Ian Mar 20 '17 at 19:47
  • $\begingroup$ Correct! I'll fix that... $\endgroup$ – R.W Mar 20 '17 at 19:48
  • $\begingroup$ But the spanning set and basis notion sometimes coincide the difference is that one has to have vectors linearly independent between each other and the spanning doesn't. The definition os a linear span is different and the questions is about in my interpretation an intuitive notion of what is a basis $\endgroup$ – R.W Mar 20 '17 at 19:56
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A basis for a vector space $V$ is a linearly independent set that spans $V$. If $V$ is given as the span of some set of vectors (as is often the case, for instance when $V$ is the image of some linear transformation), then a basis can be obtained by throwing out redundant vectors.

Note that in general, if you want to find a basis for the span of some given set of vectors, you cannot just throw out vectors willy-nilly until you have a linearly independent set. If you try that, you will probably get a linearly independent set with a smaller span than your original set of vectors. Instead, you need to actually determine which vectors are redundant before you throw any out. The way that this is done in linear algebra classes is with Gaussian elimination: a basis for the column space of $A$ is given by the columns of $A$ which become pivot columns in the echelon form of $A$.

For a toy example:

$$A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix}.$$

The columns of this matrix are clearly linearly dependent and span $\mathbb{R}^2$. But if I try to throw out the first column, then the span will shrink to just the $y$-axis. On the other hand, this matrix is already in echelon form, with the first two columns being pivot columns, so a basis for the span of these three vectors is $\left \{ \begin{bmatrix} 1 \\ 0 \end{bmatrix},\begin{bmatrix} 0 \\ 1 \end{bmatrix} \right \}$.

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You are more or less correct. One talks of a basis in respect to a linear (sub)space $S$, i.e. a set of vectors that is closed under addition and multiplication by a scalar (meaning that, for any scalar $\alpha$, and any $v_1\in S$ and $v_2\in S$, you are guaranteed that both $v_1+v_2\in S$ and $\alpha v_1\in S$).

A basis of $S$ is then a set $V$ of linearly independent vectors, such that you can obtain any non-null vector in $S$ as a linear combination of vectors of $V$ (i.e. by multiplying vectors of $V$ by some scalars and adding the results together).

Note that the requisite of linear independence is in some sense a minimality condition: if the vectors are not linearly independent, you can always discard at least one of them (but not any arbitrary one of them) and still obtain everything you could obtain before, by "simulating" the discarded vector with a linear combination of the remaining ones.

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  • $\begingroup$ So, what I mentioned in the question is not completely correct. Besides the fact that the column vectors are all linearly independent, they also need to satisfy the condition which for any scalar $\alpha$, and any $v_1\in S$ and $v_2\in S$, you are guaranteed that both $v_1+v_2\in S$ and $\alpha v_1\in S$). Is that what you mean? $\endgroup$ – Rongeegee Mar 20 '17 at 23:10

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