Thread: Can a vectorspace ever contain a 4 vector basis?

A vectorspace with 1 basis vectors is a straight line.
A vectorspace with 2 basis vectors is a plane.
A vectorspace with 3 basis vectors is a space.

So does such a thing as a 4 basis vectors exist in a vectorspace?

If a 4th vector existed, then this 4th vector could always be written as a linear combination of the 3 other vectors no matter what, and the vectors of a basis always have to be linearly independent, thus a 4th vector would make the vector set linearly dependent (since they can be written as a linear combination)... so my guess is that a basis can never be more than 3 vectors?

you are confusing vector space with 3-dimensional physical space. while 3-dimensional physical space can be modelled by a vector space (and indeed, it helps to visualize certain concepts such as orthogonal, and length of a vector by imagining things that might exist in our 3-dimensional world), the concept of vector space is MUCH more versatile than just describing physical situations.

for example, the set of all real polynomials of degree 3 or less, is certainly a valid set of things to contemplate. we can write such a polynomial as:

$\displaystyle a + bx + cx^2 + dx^3$, which we can think of as a 4-vector (a,b,c,d).

this set obeys all the axioms of a vector space over $\displaystyle \mathbb{R}$, and i claim that {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)} is a basis.

now by this, i mean the 4 polynomials:

$\displaystyle 1 + 0x + 0x^2 + 0x^3 = 1, 0 + 1x + 0x^2 + 0x^3 = x$,

$\displaystyle 0 + 0x + 1x^2 + 0x^3 = x^2, 0 + 0x + 0x^2 + 1x^3 = x^3$

it is clear that $\displaystyle \{1,x,x^2,x^3\}$ span the set of all polynomials of degree 3 or less, just use the coefficients as the scalar multiples for the linear combination of basis elements.

suppose $\displaystyle p(x) = a + bx + cx^2 + dx^3$ was the constant 0-function: 0(x) = 0, for all x (a horizontal line for its graph: the x-axis).

then p(0) = a, so a = 0. so $\displaystyle p(x) = x(b + cx + dx^2)$. since this is always 0, for all x, and x is not 0 for any x BUT 0, it must be the case for every x except 0, that:

$\displaystyle q(x) = b + cx + dx^2 = 0$, for every x, except possibly 0. but surely such a function is continuous, so it has to be 0 at x = 0 as well, by continuity.

and THAT means that q(0) = b = 0, so $\displaystyle p(x) = cx^2 + dx^3 = x^2(c + dx)$.

again, this means that c+dx = 0 for every x except possibly 0, so from x = -1, we see that:

c - d = 0, so c = d.

and from x = 1, we have:

0 = c + d = 2c, so c = d = 0.

so the only polynomial p(x), of degree 3 or less, which is 0 for ALL x, is the polynomial:

$\displaystyle 0 + 0x + 0x^2 + 0x^3$, which proves that $\displaystyle \{1,x,x^2,x^3\}$ is linearly independent, and is thus a basis.

(of course it's easy to see that if a(1,0,0,0) + b(0,1,0,0) + c(0,0,1,0) + d(0,0,0,1) = (0,0,0,0), we must have a = b = c = d = 0, but i deliberately wanted to show this another way, directly from facts you know about polynomials).

so there's one vector space which definitely has a basis of 4 elements. in fact, there's no reason to stop at 4, we can have vector spaces with 5,6,7 or any positive integer dimension, in fact there are even vector spaces with infinite bases.

now, you might think that such things are just "abstract" and don't have anything to do with "real-life stuff". but think of linear transformations in 3-dimensional space, such as rotation, or reflection (these are behaviors which real things exhibit, right?). these can be described by 3x3 matrices, and the set of all real 3x3 matrices is a vector space with 9 dimensions.

Originally Posted by MathIsOhSoHard

So does such a thing as a 4 basis vectors exist in a vectorspace?
If a 4th vector existed, then this 4th vector could always be written as a linear combination of the 3 other vectors no matter what, and the vectors of a basis always have to be linearly independent, thus a 4th vector would make the vector set linearly dependent (since they can be written as a linear combination)... so my guess is that a basis can never be more than 3 vectors?

That is completely false.
$\displaystyle \{<1,0,0,0>,<0,1,0,0>,<0,0,1,0>,<0,0,0,1>\}$ there is a set of for linearly independent vectors.
There are vector spaces of every dimension.

Every 4-dimensional vector has a basis containing four vectors. But you are correct that no 3 or fewer dimensional vector space can have a basis containing four vectors, because then it is true that one is a linear combination of the others.

Every n-dimensional vector space has a basis containing exactly n vectors. No set with fewer than n vectors is a basis as it does not span the space, and no set containing greater than n vectors is a basis because it is not linearly independent.

(Two others posted while I was typing this)

Originally Posted by AlexP

Every 4-dimensional vector has a basis containing four vectors. But you are correct that no 3 or fewer dimensional vector space can have a basis containing four vectors, because then it is true that one is a linear combination of the others.

Every n-dimensional vector space has a basis containing exactly n vectors. No set with fewer than n vectors is a basis as it does not span the space, and no set containing greater than n vectors is a basis because it is not linearly independent.

(Two others posted while I was typing this)

but it's good that the original poster should see how different people respond to this question. differing points of view often deepen our understanding of things

Thanks everybody from your inputs, it's really appreciated!
I realize that I was wrong in my assumption. However this is the root of my confusion:

"Consider a set V of vectors in space, going from origo.
Is there a subspace which has the dimensions 0, 1, 2, 3 and 4?"

I understand how a subspace 0, 1, 2, and 3 exists (zero-vector, straight line, plane and space)...
But when it comes to the subspace with the dimension 4, I get a bit confused.
Because according to my book, there exist no subspace with the dimensions 4.

But if we assume that the vectorspace V is 6 dimensions, then wouldn't i be possible for the subspace to have anywhere between 0 and 6 dimensions as well?
So why does my book say that a 4th dimension subspace cannot exist?

Originally Posted by MathIsOhSoHard

if we assume that the vectorspace V is 6 dimensions, then wouldn't i be possible for the subspace to have anywhere between 0 and 6 dimensions as well?
So why does my book say that a 4th dimension subspace cannot exist?

There are two possibilities:
1) You may have read your book incorrectly. You may have left out some conditions.
or
2) the book could be wrong.

Originally Posted by MathIsOhSoHard

Thanks everybody from your inputs, it's really appreciated!
I realize that I was wrong in my assumption. However this is the root of my confusion:

"Consider a set V of vectors in space, going from origin.
Is there a subspace which has the dimensions 0, 1, 2, 3 and 4?"

i believe what is confusing you is the ambiguous use of the word "space" here. your book should be more precise, such as:

"Consider a set V of vectors in 3-dimensional space, considered as $\displaystyle \mathbb{R}^3$. Does there exist a subspace of dimension 4?"

the answer to THAT question, which is more accurately formulated, is: no.

Originally Posted by MathIsOhSoHard

But if we assume that the vectorspace V is 6 dimensions, then wouldn't i be possible for the subspace to have anywhere between 0 and 6 dimensions as well?

Yes, that is true. And remember that the only 0-dimensional subspace is the trivial subspace containing only the zero vector, and the only 6-dimensional subspace is V itself.

There's an easy way to construct a subspace of dimension $\displaystyle k$ for $\displaystyle 1 \le k \le 5$. Find a basis for $\displaystyle V$ (which necessarily contains 6 vectors), then choose $\displaystyle k$ vectors from this basis (let's call this set $\displaystyle S$). Any subset of a linearly independent set is linear independent as well, and so the $\displaystyle k$ vectors will form a basis for the subspace $\displaystyle \mbox{span}(S)$. In fact, every subset of the basis for $\displaystyle V$ containing $\displaystyle k$ vectors will form the basis for a $\displaystyle k$-dimensional subspace. This is even true of $\displaystyle \O$, the empty set, because $\displaystyle \mbox{span}(\O)=\{0\}$, the trivial subspace.

So in summary: yes, there are subspaces of dimension $\displaystyle k$ for $\displaystyle 0 \le k \le n$, where $\displaystyle n$ is the dimension of the main vector space $\displaystyle V$, and it is easy to find some of them simply by using subsets of the basis of $\displaystyle V$.