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Determine whether the given set S is a subspace of the vector space V.

  • $V=P_5$,  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 $u+v$.

$u+v=(0,1,2,3)+(-1,0-1,0)$

$=(0+(-1),1+0,2+(-1),3+0)$

$u+v=(-1,1,1,3)$

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