Which of the following sets (with natural addition and multiplication by a scalar) are vector spaces. Justify your answer.
a) The set of all continuous functions on the interval $[0, 1]$;
b) The set of all non-negative functions on the interval $[0, 1]$;
c) The set of all polynomials of degree exactly $n$;
d) The set of all symmetric $n\times n$ matrices, i.e. the set of matrices $A =(a_{j,k})^n_{j,k=1}$ such that $A^T = A$.
Solution:
a) Yes. Let $a,b$ be complex numbers, and $f(x),g(x)$ continuous functions on the interval $[0,1]$. Then $af(x)+bg(x)$ is also continuous on $[0,1]$. Therefore, the set is a vector space.
b) No. Let $f(x)\equiv 1$. Then $f(x)$ is a non-negative function on the interval $[0,1]$. However, its additive inverse $-f(x)\equiv -1$ which is a negative function on $[0,1]$ and does not belong the set. Therefore, this set is not a vector space.
c) No. The difference of two polynomials of degree $n$ may not be of degree $n$. For example, let $f(x)=x^n+1$ and $g(x)=x^n$, then
$$
f(x)-g(x)=(x^n+1)-x^n=1.
$$
d) Yes.