MHB Linear Subspaces: Properties and Examples

[Guest2]

For the brief explanation: $\mathcal{P}$ contains $0$ by choice $p(x) = 0$ and polynomial plus a polynomial is a polynomial, and a scalar times a polynomial is a polynomial. So $\mathcal{P}$ is a non-empty subset of $\mathcal{C}^{\infty}$ that's closed under addition and scalar multiplication, and hence a subspace of $\mathcal{C}^{\infty}$.

(i) Not closed under addition. $p(x) = x^2$ and $q(x) = -x^2+x+1$ for are polynomials of precisely degree $2$ but their sum $(p+q)(x) = x+1$, a polynomial of degree $1.$

(ii) The set of even polynomials is not empty because we can take $p(x) = 0$. The sum of two even functions is even, so it's closed under addition. Also constant times an even polynomial is still an even polynomial, so it's closed under scalar multiplication. Hence it's a subspace of $\mathcal{P}$.

(iii) Let $p(x) = x^2$ and $q(x) = x^3$ then $p'(1) = 1$ and $q'(1) = 1$ but $(p+q)'(1) = 2$, so it's closed under addition.

(iv) It's not empty since it contains $0$ by the choice $\lambda = \mu = 0$. Let $p(x) = \lambda x + \mu x^2+ (\lambda-\mu)x^3$ and $q(x) = \lambda' x + \mu' x^2+ (\lambda'-\mu')x^3$ then $(p+q)(x) = (\lambda+\lambda') x + (\mu+\mu') x^2+ (\lambda-\mu+\lambda'-\mu')x^3$ which is again in the set. Also multiplying by a constant, we're still in the set, so it's a subspace.

Is this alright, and how do I do part (b)?

 

[Guest2]

(b) (i) Not linear. Let $f(x) = x$ then $F_1(2f(x)) = (2x)^2 = 4x^2 \ne 2F_{1}(f(x)) = 2f(x) = 2x^2$

(ii) Linear. Let $f(x) \mapsto f(x)+g(x)$ then $$F_2(f(x)+g(x)) = x(f(x)+g(x)) = xf(x)+xg(x) = F_{2}(f(x))+F_2(g(x)).$$ Also if $\lambda \in \mathbb{R}$ then $F_2(\lambda f(x)) = \lambda x f(x) = \lambda F_2(f(x))$

(iii) Linear. Let $f(x) \mapsto f(x)+g(x)$ then $$F_3(f(x)+g(x)) = x^2(f(x)+g(x))''-2x(f(x)+g(x))' =x^2f''(x)-2xf'(x)+g''(x)-2xg'(x) = F_3(f(x))+F_3(g(x)).$$ Also if $\lambda \in \mathbb{R}$ then $F_3(\lambda f(x)) = \lambda x^2 f''(x) -2\lambda xf'(x)=\lambda (x^2 f''(x) -2 xf'(x)) = \lambda F_3(f(x))$

(iv) Not linear. $F_{4}(0) = x$ while $0 F_4(f(x)) = 0$.

How do I find the null space of:

1. $F_2(f(x)) = xf(x)$

2. $F_3(f(x)) = x^2f''(x)-2xf'(x)$

What's the matrix associated with these spaces?

 

[Deveno]

Linear transformations on an infinite-dimensional vector space don't generally have "matrices", because matrix multiplication would involve infinite sums, and such things are (generally) undefined.

But that doesn't stop us from findng the null space, which is (in this case), for each transformation $F$ the set of polynomials:

$\{p(x): F(p(x) \equiv 0\}$, where the $0$ is the $0$-function: $0(t) = 0$, for all $t \in \Bbb R$ (this is a constant function).

If $F(p(x)) = xp(x)$, then we must have:

$tp(t) = 0$, for ALL real $t$.

If $t \neq 0$, then the fact that $\Bbb R$ is a field, forces $p(t) = 0$ for ALL $t \neq 0$.

By continuity of polynomials, this also forces $p(0) = 0$, so we conclude $p$ MUST be the $0$-polynomial (that is, the $0$-function).

Another way to see this, is by regarding a polynomial as a finite sequence of its coefficients: if

$p(x) = a_0 + a_1x +\cdots + a_nx^n$ we can regard this as the sequence:

$(a_0,a_1,\dots,a_n)$ (we can extend this to an infinite sequence by appending infinitely many 0's at the end).

Then $F$ does this on infinite sequences:

$(a_0,a_1,\dots,a_n,0,\dots) \mapsto (0,a_0,a_1,\dots,a_n,0,\dots)$

If the image is all 0's, we must have started with all 0's, for any non-zero entry in the domain sequence would be mapped to a non-zero entry in the image sequence (just "shifted up one space to the right").

Your second example will be more challenging:

We can factor out an $x$ to get:

$F(f(x)) = x(xf''(x) - 2f'(x))$, since $x$ is not identically 0, for the result to be identically 0, we need:

$xf''(x) = 2f'(x)$.

My suggestion:

Write $f(x) = a_0 + a_1x +\cdots + a_nx^n$, and compute $f'(x)$ and $f''(x)$. It might hep to start with low values of $n$.