Subspace spanned by subsets of polynomials

[Guest2]

In the linear space of all real polynomials $p(t)$, describe the subspace spanned by each of the following subsets of polynomials and determine the dimension of this subspace.

(a) \(\displaystyle \left\{1,t^2,t^4\right\}\), (b)\(\displaystyle \left\{t,t^3,t^4\right\}\), (c) \(\displaystyle \left\{t,t^2\right\}\), (d) $\left\{1+t, (1+t)^2\right\}$

(a) Let $a_1, a_2, a_3$ be scalars. Then the subspace spanned by \(\displaystyle \left\{1,t^2,t^4\right\}\) is the set $ l(t) = \left\{a_1t^4+a_2t^2+a_3\right\}$. This is the space of all biquadratic polynomials. To find the dimension, we need to show that \(\displaystyle \left\{1,t^2,t^4\right\}\) is a basis for this set. We already have that it spans set of all biquadratic polynomials. To find that it's linearly independent, suppose that $a_1t^4+a_2t^2+a_3 = 0$ for all $t$. But this can only happen if $a_1 = a_2 = a_3 = 0$. Therefore \(\displaystyle \left\{1,t^2,t^4\right\}\) is a basis for $l(t)$. Therefore $\text{dim}(l(t)) = 3$. Is this close enough?

(b) I think \(\displaystyle \left\{t,t^3,t^4\right\}\) spans the set $ l(t) = \left\{a_1t+a_2t^3+a_3t^4: a_1, a_2, a_3 \in \mathbb{R}\right\}$. It's the space of all polynomials of degree less than or equal to $4$ with no quadratic and constant terms. \(\displaystyle \left\{t,t^3,t^4\right\}\) is also a basis for this space because $a_1t+a_2t^3+a_3t^4 = 0 $ for all $t$ implies $a_1 = a_2 = a_3 = 0$. Therefore $\text{dim}(l(t)) = 3$.

(c) \(\displaystyle \left\{t,t^2\right\}\) spans the set $ l(t) = \left\{a_1t+a_2t^2: a_1, a_2 \in \mathbb{R}\right\}$. It's the space of all polynomials of degree less than or equal to $2$ with no constant terms. \(\displaystyle \left\{t, t^2\right\}\) is also a basis for this space because $a_1t+a_2t^2= 0 $ for all $t$ implies $a_1 = a_2 = 0$. Therefore $\text{dim}(l(t)) = 2$.

(d) $\left\{1+t,(1+t)^2\right\}$ spans the set $ l(t) = \left\{a_1(1+t)+a_2(1+t)^2: a_1, a_2 \in \mathbb{R}\right\}$. It's the space of all quadratic functions. \(\displaystyle \left\{1+t, (1+t)^2\right\}\) is also a basis for this space because $a_1(1+t)+a_2(1+t)^2 = (a_1+a_2)+(2a_2+a_1)t+a_2t^2 = 0 $ for all $t$ implies $a_1 = a_2 = 0$. Therefore $\text{dim}(l(t)) = 2$.

 

[Evgeny.Makarov]

(a) Let $a_1, a_2, a_3$ be scalars. Then the subspace spanned by \(\displaystyle \left\{1,t^2,t^4\right\}\) is the set $ l(t) = \left\{a_1t^4+a_2t^2+a_3\right\}$.

It should say $\{a_1t^4+a_2t^2+a_3\mid a_1,a_2,a_3\in\Bbb R\}$.

This is the space of all biquadratic polynomials.

Depending on the definition, "biquadratic polynomial" may mean a polynomial of degree 4 (with other conditions). This subspace also contains polynomials of lesser degree.

(d) $\left\{1+t,(1+t)^2\right\}$ spans the set $ l(t) = \left\{a_1(1+t)+a_2(1+t)^2: a_1, a_2 \in \mathbb{R}\right\}$. It's the space of all quadratic functions.

The dimension of the space over $\Bbb R$ of polynomials of degree $\le2$ is $3$, so it cannot be spanned by two polynomials.

 

[Guest2]

Thank you.

It should say $\{a_1t^4+a_2t^2+a_3\mid a_1,a_2,a_3\in\Bbb R\}$.

Depending on the definition, "biquadratic polynomial" may mean a polynomial of degree 4 (with other conditions). This subspace also contains polynomials of lesser degree.

So the space of all even polynomials of degree $ \le 4$ would have been a better description?

The dimension of the space over $\Bbb R$ of polynomials of degree $\le2$ is $3$, so it cannot be spanned by two polynomials.

How would you describe the space $l(t) = \left\{a_1(1+t)+a_2(1+t)^2: a_1, a_2 \in \mathbb{R}\right\}$?

 

[Evgeny.Makarov]

How would you describe the space $l(t) = \left\{a_1(1+t)+a_2(1+t)^2: a_1, a_2 \in \mathbb{R}\right\}$?

You can provide different bases, such as $(1+t,1+2t+t^2)$, $(t+t^2,1+2t+t^2)$ or $(1+t,t+t^2)$.