Let $P_n$ be the set of polynomials of degree at most $n$ with real coefficients. Is the set of polynomials of the form $p(t) = a + t^2,$ such that '$a$' is an element of $\mathbb{R},$ a subspace of $P_n?$
I know I need to check if it contains the zero vector first. How would I do this? I have two ideas:
$1.$ plug $0$ into '$a$' and have a function $g(t) = t^2$ then add it to $p(t)$ to get $p(t) + g(t) = a + 2t^2$ which is not in the form, or
$2.$ plug $0$ into '$a$' and also for the coefficient of $t^2?$
Thanks!
$\endgroup$ 23 Answers
$\begingroup$Let $V$ be a vector space and $W\subset V$. We want $W$ to satisfy three key axioms for it to fit the definition of subspace.
$1$. $\{0\}\in W$
$2$. $w_1+w_2=w_3\in W$ (closure under vector addition)
$3$. $cw\in W$ (closure under scalar multiplication)
For the subset of polynomials $W$ defined by $p(t)=a+t^2$, we don't have closure under addition, because we have $p(t)+q(t)=(a+b)+2t^2$, which is not of the desired form.
also, the set fails closure under scalar multiplication as well, since $cp(t)=c(a+t^2)=ca+ct^2$. The only exception is $c=1$, but $W$ still fails the vector addition axiom so it is not a subspace.
Finally, the zero vector (and for polynomials, the zero polynomial -that whose all coefficients $a_0, a_1,..., a_n = 0$, and in this case, only $a = 0$) is also not in the subset except for the single case where $t=0$.
$\endgroup$ $\begingroup$Both of your two ideas work and are indeed based on the fact that the polynomial set of that form is not closed under addition(obvious, for $t^2+t^2=2t^2$). Hence it's not a subspace.
$\endgroup$ 2 $\begingroup$The zero element $P_n$ is not in the set, but I feel some clarification to the above statements is necessary: The zero of this space is the polynomial with all zero coefficients - not the polynomial that equals zero when you set $t=0$. Thus, multiplication by the zero polynomial will result in zero for all vectors (i.e., other polynomials) in this space. You never really plug in a number for $t$ when dealing with questions on this space, the vectors are the polynomial equations themselves. This is enough to conclude it is not a vector space, but it's also true that the set is not closed under vector addition or scalar multiplication as well, as outlined above.
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