# Can a basis for an $n$ dimensional vector space consist of $n+1$ basis vectors?

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.

• Isn’t dimension defined to be the cardinality of any basis?
– MPW
Mar 31, 2019 at 12:06
• The polynomials with degree at most $3$ form a space with dimension $4$ Mar 31, 2019 at 12:07
• 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. Mar 31, 2019 at 12:16
• Polynomials of degree $n$ are not a vector space. Only polynomials of degree $\le n$. Mar 31, 2019 at 12:42

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 $$V$$.