- $V=P_5$, and $S$ is the subset of $P_5$ consisting of the polynomials satisfying $p(1)>p(0)$.
- $V=R_3$, and $S$ is the set of vectors $(x_1,x_2,x_3)$ in $V$ satisfying $x_1-6x_2+x_3=5$.
- $V=R^n$ and $S$ is a set of solutions to the homogeneous linear system $Ax=0$, where $A$ is a fixed $m\times n$ matrix.
- $V=C^2(I)$, and $S$ is the subset of $V$ consisting of those functions satisfying the differential equation $y^{\prime\prime}-4y’+3y=0$.
- $V$ is the vector space of all the real-valued functions defined on the interval $[a,b]$, and $S$ is a subset of $V$ consisting of those functions satisfying $f(a)=5$.
- $V=P_n$, and $S$ is the subset of $P_n$ consisting of those polynomials satisfying $p(0)=0$.
- $V=M_n(R)$, and $S$ is the subset of all symmetric matrices.
The goal of this question is to sort out whether the given set $S$ is a subspace of the vector space $V$.
A vector space $V$ satisfies the closure property with respect to multiplication and addition as well as the distributive and associative procedure of vector multiplication by scalars. More generally, a vector space is composed of a set of vectors $(V)$, a scalar field $(F)$ along with vector addition, and scalar multiplication.
A subspace is a vector space that is contained within a larger vector space. As a result, the closure property with respect to multiplication and addition holds for a subspace as well.
Mathematically, assume that $V$ and $U$ are two vector spaces with the same definitions of vector addition and scalar multiplication, and $U$ is a subset of $V$ i.e., $U\subseteq V$, then $U$ is said to be a subspace of $V$.
Expert Answer
- We know that, a subset $S$ will be a subspace of $V$ iff for all $\alpha,\beta\in R$ and $x,y\in S$, $\alpha x+\beta y\in S$.
So $S$ will not be a subspace of $V=P_5$.
Reason
Consider two functions:
$p(x)=x^2+5$ and $q(x)=x^2-5$
$p(1)=6$ and $p(0)=5$ $\implies p(1)>p(0)$
$q(1)=-4$ and $q(0)=-5$ $\implies q(1)>q(0)$
$\implies p(x),\,q(x)\in S$
Assume that $R(x)=p(x)-2q(x)$
$R(1)=p(1)-2q(1)=6+8=14$
$R(0)=p(0)-2q(0)=5+10=15$
Hence, $R(1)<R(0)$
Therefore, $S$ is not a subspace of $P_5$.
- $S$ is not a subspace of $V=R_3$.
Reason
Let $(-1,-1,0)\in S$ so, $(-1)-(-1)6+0=-1+6=5$
Suppose that $-1(-1,-1,0)=(1,1,0)$
So that, $1-6+0=-5\neq 5$
$\implies (1,1,0)\notin S$
Therefore, $S$ is not a subspace of $R_3$.
- $S$ is a subspace of $V=R^n$
Reason
Let $x,y\in S$ then we have $Ax=0$ and $Ay=0$.
$A(\alpha x+\beta y)=\alpha Ax+\beta Ay$
$=\alpha(0)+\beta(0)=0$
$\implies \alpha x+\beta y\in S$ and hence $S$ is a subspace of $V=R^n$.
- $S$ is a subspace of $V=C^2(I)$
Reason
Let $x,y\in S$ then $x^{\prime\prime}-4x’+3x=0$ and $y^{\prime\prime}-4y’+3y=0$.
Now, $(\alpha x+\beta y)^{\prime\prime}-4(\alpha x+\beta y)’+3(\alpha x+\beta y)$
$=\alpha x^{\prime\prime}+\beta y^{\prime\prime}-4\alpha x’-4\beta y’+3\alpha x+3\beta y$
$=\alpha (x^{\prime\prime}-4x’+3x)+\beta(y^{\prime\prime}-4y’+3y)$
$=\alpha (0)+\beta (0)$
$=0$
$\implies \alpha x+\beta y\in S$ and hence $S$ is a subspace of $V=C^2(I)$.
- $S$ is not a subspace of $V$
Reason
Suppose that $f,g\in S$, then $f(a)=5$ and $g(a)=5$
$\alpha f(a)+\beta g(a)=5\alpha+5\beta$
Assume that $\alpha=1$ and $\beta=-1$
$\implies \alpha f(a)+\beta g(a)=5-5=0\notin S$
$\implies \alpha f(a)+\beta g(a)\notin S$
Therefore, $S$ is not a subspace of $V$.
- $S$ is a subspace of $V=P_n$.
Reason
Assume that $p,q\in S$, then $p(0)=0$ and $q(0)=0$
And $\alpha p+\beta q=\alpha (0)+\beta (0)=0$
$\implies \alpha p+\beta q\in S$
Therefore, $S$ is a subspace of $V=P_n$.
- $S$ is a subspace $V=M_n(R)$
Reason
Let $A,B\in S$, then $A^T=A$ and $B^T=B$
Now, $(\alpha A+\beta B)^T=(\alpha A)^T+(\beta B)^T$
$=\alpha A^T+\beta B^T=\alpha A+\beta B$
$\implies \alpha A+\beta B\in S$
Therefore, $S$ is a subspace of $V=M_n(R)$.
Example
Let $E^n$ be the Euclidean space. Suppose that $u=(0,1,2,3)$ and $v=(-1,0-1,0)$ in $E^4$. Find $u+v$.
$u+v=(0,1,2,3)+(-1,0-1,0)$
$=(0+(-1),1+0,2+(-1),3+0)$
$u+v=(-1,1,1,3)$