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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.

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.

Quote:

Originally Posted by

**Elusive1324**

Statements-

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

$\displaystyle C^1(\mathbb{R})$ is the set of differentiable functions with continuous derivatives. The subset $\displaystyle S$ is the set of all such functions with non-negative derivative at 0. To see if $\displaystyle S$ is closed under vector addition, check that $\displaystyle f,g\in S \implies f+g \in S$. This is the same as asking whether $\displaystyle f'(0),g'(0)\geq 0 \implies (f+g)'(0)\geq 0$. Similarly for the scalar multiplication case, check if $\displaystyle f'(0)\geq 0 \Rightarrow (cf)'(0)\geq 0$ for any scalar c. HINT: what if $\displaystyle c$ is negative and $\displaystyle f'(0)>0$?

Quote:

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.

Quote:

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 $\displaystyle a$ twice for different things when calculating (2).

Quote:

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.

Quote:

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

Here, $\displaystyle C^2(\mathbb{R})$ is the set of twice differentiable functions with continuous first and second derivatives. The idea is basically the same as in A.

Quote:

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.