stop2dance3l

2021-12-17

Let W be the set of all vectors p(t) in P such that $p\left(0\right)=0$. Show that W is a subspace of P and find a basis for W.

Thomas White

Expert

Solution:
W be the set of all vectors p(t) in ${P}_{2}$ such that $p\left(0\right)=0$
${P}_{2}\left(t\right)=\left\{{a}_{0}+{a}_{1}t+{a}_{2}\frac{{t}^{2}}{{a}_{0}},{a}_{1},{a}_{2}\in R\right\}$
now, $W=\left\{p\left(t\right)={a}_{0}+{a}_{1}t+{a}_{2}\frac{{t}^{2}}{p}\left(0\right)=0\right\}$
$p\left(0\right)=0$
${a}_{0}+{a}_{1}\left(0\right)+{a}_{2}\left(0\right)=0$
${a}_{0}=0$
$W=\left\{{a}_{1}t+{a}_{2}\frac{{t}^{2}}{{a}_{1}},{a}_{2}\in R\right\}$
check subspace condition,
1) let
$x+y=\left({a}_{1}t+{a}_{2}{t}^{2}\right)+\left({b}_{1}t+{b}_{2}{t}^{2}\right)$
$=\left({a}_{1}+{b}_{1}\right)t+\left({a}_{2}+{b}_{2}\right){t}^{2}$
$x+y\in W$
2) be a scalar.
$ax=a\left({a}_{1}t+{a}_{2}{t}^{2}\right)$
$=\left(a{a}_{1}\right)t+\left(a{a}_{2}\right){t}^{2}$
$ax\in W$
3) put ${a}_{1}={a}_{2}=0$
we get, $p\left(t\right)=0$
W has zero vector
Therefore W is a subspace of ${P}_{2}$
now, $W=\left\{{a}_{1}t+{a}_{2}\frac{{t}^{2}}{{a}_{1}},{a}_{2}\in R\right\}$