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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) $$\left\{1,t^2,t^4\right\}$$, (b)$$ \left\{t,t^3,t^4\right\}$$, (c) $$\left\{t,t^2\right\}$$, (d) $\left\{1+t, (1+t)^2\right\}$
(a) Let $a_1, a_2, a_3$ be scalars. Then the subspace spanned by $$\left\{1,t^2,t^4\right\}$$ 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 $$\left\{1,t^2,t^4\right\}$$ 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 $$\left\{1,t^2,t^4\right\}$$ is a basis for $l(t)$. Therefore $\text{dim}(l(t)) = 3$. Is this close enough?
(b) I think $$\left\{t,t^3,t^4\right\}$$ 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. $$\left\{t,t^3,t^4\right\}$$ 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) $$\left\{t,t^2\right\}$$ 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. $$\left\{t, t^2\right\}$$ 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. $$\left\{1+t, (1+t)^2\right\}$$ 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 $\text{dim}(l(t)) = 2$.
(a) $$\left\{1,t^2,t^4\right\}$$, (b)$$ \left\{t,t^3,t^4\right\}$$, (c) $$\left\{t,t^2\right\}$$, (d) $\left\{1+t, (1+t)^2\right\}$
(a) Let $a_1, a_2, a_3$ be scalars. Then the subspace spanned by $$\left\{1,t^2,t^4\right\}$$ 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 $$\left\{1,t^2,t^4\right\}$$ 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 $$\left\{1,t^2,t^4\right\}$$ is a basis for $l(t)$. Therefore $\text{dim}(l(t)) = 3$. Is this close enough?
(b) I think $$\left\{t,t^3,t^4\right\}$$ 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. $$\left\{t,t^3,t^4\right\}$$ 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) $$\left\{t,t^2\right\}$$ 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. $$\left\{t, t^2\right\}$$ 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. $$\left\{1+t, (1+t)^2\right\}$$ 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 $\text{dim}(l(t)) = 2$.
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