# Math Help - Determine whether a set S is a subspace of a vector space V

1. ## Determine whether a set S is a subspace of a vector space V

The exercise is attached. Each statement defines a subspace S and a linear space V. I'm tasked to determine whether S is a subspace of V.

Definition of subspace:
A subset W of of a linear space V is called a subspace of V if
(1) The subset contains the neutral element 0 of V
(2) W is closed under addition
(3) W is closed under multiplication

My attempt:

Statements-
A. I don't understand the notation nor how this statement defines S

B. Being a subset of V, set S contains all real-defined functions defined on [a,b]. Also if f(a)=0 and f(b)=0, then S contains the neutral element of V. If we have some f(a)=f(b), multiplying both sides of the equation by the same scalar preserves the equality. (1),(2),(3) is satisfied so it is a subspace of V.

C. p(0)=0 is the neutral element of V, and for polynomials in S, p(0)=0. If s(x)=ax^3+bx and t(x)=cx^3+dx, then (s+t)(x)=(a+c)x^3+(b+d)x which has the form p(x)=ax^3+bx so (2) is satisfied. If s(x)=ax^3+bx and k is a scalar, k(s(x)) = (ka)x^3 + (kb)x, which retains the form so (3) is satisfied. (1),(2),(3) is satisfied so it is a subspace of V.

D. The 0 square matrix is symmetric and in V. For symmetric square matrices A and B, taking A+B by taking the sum of corresponding entries will give a symmetric matrix. Likewise if a symmetric matrix is multiplied by a scalar, its entries will still be symmetric. (1),(2),(3) is satisfied so it is a subspace of V.

E. vector v=(0,0,0) is the neutral element in V but it does not satisfy the conditions to be in set S. S here is not a subspace.

F. I don't understand the notation for how they defined V.

G. For the set of all polynomials of any degree f(x)=0 is the zero element. For all polynomials, f(0)=0. Set is is closed under addition but not multiplication. p(1)> 0 => p(0) but if you multiply the inequality by a negative scalar, the inequality doesn't hold. Not a subspace of V.

2. ## Re: Determine whether a set S is a subspace of a vector space V

I will give a detailed hint for the first one, and address your concerns for the rest.

Originally Posted by Elusive1324

Statements-
A. I don't understand the notation nor how this statement defines S
$C^1(\mathbb{R})$ is the set of differentiable functions with continuous derivatives. The subset $S$ is the set of all such functions with non-negative derivative at 0. To see if $S$ is closed under vector addition, check that $f,g\in S \implies f+g \in S$. This is the same as asking whether $f'(0),g'(0)\geq 0 \implies (f+g)'(0)\geq 0$. Similarly for the scalar multiplication case, check if $f'(0)\geq 0 \Rightarrow (cf)'(0)\geq 0$ for any scalar c. HINT: what if $c$ is negative and $f'(0)>0$?

B. Being a subset of V, set S contains all real-defined functions defined on [a,b]. Also if f(a)=0 and f(b)=0, then S contains the neutral element of V. If we have some f(a)=f(b), multiplying both sides of the equation by the same scalar preserves the equality. (1),(2),(3) is satisfied so it is a subspace of V.
You didn't check if S is closed under vector addition.

C. p(0)=0 is the neutral element of V, and for polynomials in S, p(0)=0. If s(x)=ax^3+bx and t(x)=cx^3+dx, then (s+t)(x)=(a+c)x^3+(b+d)x which has the form p(x)=ax^3+bx so (2) is satisfied. If s(x)=ax^3+bx and k is a scalar, k(s(x)) = (ka)x^3 + (kb)x, which retains the form so (3) is satisfied. (1),(2),(3) is satisfied so it is a subspace of V.
OK, although you used the variable $a$ twice for different things when calculating (2).

D. The 0 square matrix is symmetric and in V. For symmetric square matrices A and B, taking A+B by taking the sum of corresponding entries will give a symmetric matrix. Likewise if a symmetric matrix is multiplied by a scalar, its entries will still be symmetric. (1),(2),(3) is satisfied so it is a subspace of V.

E. vector v=(0,0,0) is the neutral element in V but it does not satisfy the conditions to be in set S. S here is not a subspace.
yes.

F. I don't understand the notation for how they defined V.
Here, $C^2(\mathbb{R})$ is the set of twice differentiable functions with continuous first and second derivatives. The idea is basically the same as in A.

G. For the set of all polynomials of any degree f(x)=0 is the zero element. For all polynomials, f(0)=0. Set is is closed under addition but not multiplication. p(1)> 0 => p(0) but if you multiply the inequality by a negative scalar, the inequality doesn't hold. Not a subspace of V.
ok.