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Give an example of a $3$-dimensional subspace of $P_4$ which contains the polynomials $3$+$2t^2$, $t^4$, and $1+2t+3t^3$.

I have no idea how to even get started with this problem. Can I get some help? (Studying)

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If the three polynomials $p_1(t)= 3+2 t^2$, $p_2(t) = t^4$ and $p_3(t) = 1 + 2t + 3 t^3$ are linearly independent, then the can be used as a basis to form a 3 dimensional subspace.

To show that they are linearly independent, I need to show that if $\sum_{k=1}^3 \alpha_k p_k = 0$, then the $\alpha_k = 0$. This gives $\sum_{k=1}^3 \alpha_k p_k(t) = (3 \alpha_1+\alpha_2)+2 \alpha_2 t + 2 \alpha_1 t^2+3 \alpha_2 t^3 + \alpha_3 t^4 = 0$. If we differentiate 4 times, we get that $\alpha_3 = 0$. If we differentiate 3 times and set $t=0$, we get that $\alpha_2 = 0$. Just setting $t=0$ gives $\alpha_1=0$, hence these three polynomials are linearly independent.

So, a three dimensional space that contains the given polynomials is given by $S = \text{sp} \{p_k\}_{k=1}^3 = \{ \sum_{k=1}^3 \alpha_k p_k \}_{\alpha \in \mathbb{R}^3}$, where $p_1,p_2,p_3$ are given by above.

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    Well, no, the summation convention you refer to is just a convention to avoid a proliferation of $\sum$ symbols, but that is not an issue here. The dimension of a subspace is the maximum number of linearly independent vectors that you can find in that space. A set of vectors $v_k$ is linearly independent iff the only solution to $\sum_k \alpha_k v_k = 0$ is $\alpha_k = 0$.2012-11-07
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Perhaps $\text{span}\{ 3+2t^2,t^4,1+2t+3t^3 \}$? I suppose you need to argue that the set in question is indeed linearly independent, but that shouldn't be hard.

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    @LearningPython The question is a bit strange. It seems too simple. Maybe you are supposed to see this as a subspace for some other reason? The kernel of some map, the image of another? Any span is a subspace so my answer is true, but maybe there is a more interesting one...2012-11-07