Can a basis for an $n$ dimensional vector space consist of $n+1$ basis vectors?
I read that the set $\{x^3-2,x+x^2,1,x\}$ is a basis for the vector space polynomials of degree $3$ but I'm not exactly convinced.
$\begingroup$
$\endgroup$
4
-
1$\begingroup$ Isn’t dimension defined to be the cardinality of any basis? $\endgroup$– MPWMar 31, 2019 at 12:06
-
1$\begingroup$ The polynomials with degree at most $3$ form a space with dimension $4$ $\endgroup$– PeterMar 31, 2019 at 12:07
-
$\begingroup$ It's a theorem, that if a vector space has a basis of $n$ vectors, then any basis of that space must also have $n$ vectors. This allows the unambiguous definition of $n$ as the "dimension" of that space. $\endgroup$– SimonMar 31, 2019 at 12:16
-
$\begingroup$ Polynomials of degree $n$ are not a vector space. Only polynomials of degree $\le n$. $\endgroup$– BernardMar 31, 2019 at 12:42
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
The cardinality of a basis is always equal to the dimension of the space, so a basis of a $n-$dimensional space has exactly $n$ elements.
Note that the space of polynomial of degree $\leq 3$ is a space of dimension $4$ (its canonical basis is $(1, X, X^2, X^3)$).
answered Mar 31, 2019 at 12:09
$\begingroup$
$\endgroup$
A set of $\ n+1\ $ vectors in a $\ n\ $-dimensional vector space is always linear dependent, hence cannot form a base.
$\begingroup$
$\endgroup$
By definiton, an $n$ dimensional vector space $V$ means that any basis of $V$ has cardinality $n$. So, an $n$-dimensional vector space cannot have a basis of cardinality $n+1$. In your case, the vector space you are looking at has a basis of cardinality $4$ ,i.e., $\{1,x,x^2,x^3\}$ and so all the bases of the vector space of all polynomials of degree $\leq 3$ have cardinality $4$.
To convince you that the set $\{x^{3}-2,x+x^{2},1,x\}$ is indeed a basis, it is sufficient to check that the set is linearly independent.
To show this, let $a_{0}(x^{3}-2)+a_{1}(x+x^{2})+a_{2}(1)+a_{3}x=0$ and prove that the scalars $a_{0}=a_{1}=a_{2}=a_{3}=0$.
PS: I am also assuming that the coefficients of the polynomials are in some ring.
$\begingroup$
$\endgroup$
The problem is to understand what exactly the dimension of a vector space is. I'll recall this result, which is the basis for the notion of dimension:
Let $V$ be a finite-dimensional space. All bases of $V$ have the same number of elements. Furthermore, the following numbers are the same:
- The minimal number of elements in a generating family for $V$.
- The maximal number of elements in a linearly independent family of vectors in $V$.
- The number of elements in a basis for $V$.
