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.

  • 1
    $\begingroup$ Isn’t dimension defined to be the cardinality of any basis? $\endgroup$
    – MPW
    Mar 31, 2019 at 12:06
  • 1
    $\begingroup$ The polynomials with degree at most $3$ form a space with dimension $4$ $\endgroup$
    – Peter
    Mar 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. $\endgroup$
    – Simon
    Mar 31, 2019 at 12:16
  • $\begingroup$ Polynomials of degree $n$ are not a vector space. Only polynomials of degree $\le n$. $\endgroup$
    – Bernard
    Mar 31, 2019 at 12:42

4 Answers 4


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 $(1, X, X^2, X^3)$).


A set of $\ n+1\ $ vectors in a $\ n\ $-dimensional vector space is always linear dependent, hence cannot form a base.


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.


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:

  1. The minimal number of elements in a generating family for $V$.
  2. The maximal number of elements in a linearly independent family of vectors in $V$.
  3. The number of elements in a basis for $V$.

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