# Determine whether the given set S is a subspace of the vector space V. A. V=P_5, and S is the subset of P_5 consisting of those polynomials satisfying

Determine whether the given set S is a subspace of the vector space V.
A. V=${P}_{5}$, and S is the subset of ${P}_{5}$ consisting of those polynomials satisfying p(1)>p(0).
B. $V={R}_{3}$, and S is the set of vectors $\left({x}_{1},{x}_{2},{x}_{3}\right)$ in V satisfying ${x}_{1}-6{x}_{2}+{x}_{3}=5$.
C. $V={R}^{n}$, and S is the set of solutions to the homogeneous linear system Ax=0 where A is a fixed m×n matrix.
D. V=${C}^{2}\left(I\right)$, and S is the subset of V consisting of those functions satisfying the differential equation y″−4y′+3y=0.
E. V is the vector space of all real-valued functions defined on the interval [a,b], and S is the subset of V consisting of those functions satisfying f(a)=5.
F. V=${P}_{n}$, and S is the subset of ${P}_{n}$ consisting of those polynomials satisfying p(0)=0.
G. $V={M}_{n}\left(R\right)$, and S is the subset of all symmetric matrices
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Solution.

A subset S is a subspace of V if $\alpha x+\beta y\in S$ for all x, $y\in S$ and $\alpha ,\beta \in R$

A) Not a subspace

and $p\left(1\right)>p\left(0\right)$

and $q\left(1\right)>q\left(0\right)$

$⇒p\left(x\right),q\left(x\right)\in S$

Consider $R\left(x\right)=p\left(x\right)-2q\left(x\right)$

$R\left(1\right)=p\left(1\right)-2q\left(1\right)=6+8=24$

$R\left(0\right)=p\left(0\right)-2q\left(0\right)=5+10=25$

Here $R\left(1\right)

$⇒R\left(x\right)\ne \in S$

$⇒S$ is not a subspace

B) Not a subspace

$\left(-1,-1,0\right)\in S$ as $\left(-1\right)-\left(-1\right)6+0=-1+6=5$

Consider $-1\left(-1,-1,0\right)=\left(1,1,0\right)$

$1-6+0=-5\ne 5$

$⇒\left(1,1,0\right)\ne \in S$

$⇒S$ is not subspace

C) Subspace

Let $x,y\in S$ then $Ax=0$ and $Ay=0$

$A\left(\alpha x+\beta y\right)=\alpha Ax+\beta Ay$

$=\alpha \left(0\right)+\beta \left(0\right)$

$=0$

$⇒\alpha x+\beta y\in S$

$⇒S$ is a subspace

D) Subspace

Let $x,y\in S$ then ${x}^{″}-4{x}^{\prime }+3x=0$

and ${y}^{″}-4{y}^{\prime }+3y=0$

Consider $\alpha x+\beta y$

$\left(\alpha x+\beta y{\right)}^{″}-4\left(\alpha x+\beta y{\right)}^{\prime }+3\left(\alpha x+\beta y\right)$

$=\alpha {x}^{″}+\beta {y}^{″}-4\alpha {x}^{\prime }-4\beta {y}^{\prime }+3\alpha x+3\beta y$

$=\alpha \left({x}^{″}-4{x}^{\prime }+3x\right)+\beta \left({y}^{″}-4{y}^{\prime }+3y\right)$

$=\alpha \left(0\right)+\beta \left(0\right)$

$=0$

$⇒\alpha x+\beta y\in S$

$⇒S$ is a subspace

E) Not a subspace

Let $f,g\in S$ then $f\left(a\right)=5$ and $g\left(a\right)=5$

$\alpha f\left(a\right)+\beta g\left(a\right)=5\alpha +5\beta$

Let $\alpha =1$ and $\beta =-1$

$⇒\alpha f\left(a\right)+\beta g\left(a\right)=5-5=0$

$⇒\alpha f\left(a\right)+Bg\left(a\right)\ne \in S$

$⇒S$ is not a subspace

F) Subspace

Let $p,g\ne \in S$ then $p\left(0\right)=0$ and $g\left(0\right)=0$

Consider $\alpha p+\beta g$