stop2dance3l

Answered

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.

Answer & Explanation

Thomas White

Expert

2021-12-18Added 40 answers

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)=\{{a}_{0}+{a}_{1}t+{a}_{2}\frac{{t}^{2}}{{a}_{0}},{a}_{1},{a}_{2}\in R\$

now,$W=\{p\left(t\right)={a}_{0}+{a}_{1}t+{a}_{2}\frac{{t}^{2}}{p}\left(0\right)=0\}$

$p\left(0\right)=0$

${a}_{0}+{a}_{1}\left(0\right)+{a}_{2}\left(0\right)=0$

${a}_{0}=0$

$W=\{{a}_{1}t+{a}_{2}\frac{{t}^{2}}{{a}_{1}},{a}_{2}\in R\}$

check subspace condition,

1) let$x={a}_{1}t+{a}_{2}{t}^{2}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}y={b}_{1}t+{b}_{2}{t}^{2};x,y\in W$

$x+y=({a}_{1}t+{a}_{2}{t}^{2})+({b}_{1}t+{b}_{2}{t}^{2})$

$=({a}_{1}+{b}_{1})t+({a}_{2}+{b}_{2}){t}^{2}$

$x+y\in W$

2)$x={a}_{1}t+{a}_{2}{t}^{2}\in W\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}a\in F$ be a scalar.

$ax=a({a}_{1}t+{a}_{2}{t}^{2})$

$=\left(a{a}_{1}\right)t+\left(a{a}_{2}\right){t}^{2}$

$ax\in W$

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,$W=\{{a}_{1}t+{a}_{2}\frac{{t}^{2}}{{a}_{1}},{a}_{2}\in R\}$

W be the set of all vectors p(t) in

now,

check subspace condition,

1) let

2)

3) put

we get,

W has zero vector

Therefore W is a subspace of

now,