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I'm very confused because we haven't got vectors or matrices here. So I really have no idea how to solve that task. I thought about converting these to one matrix somehow but it doesn't seem to work. This is no homework, it's a task from an old exam.

Let $W= \text{span}(p_{1},p_{2},p_{3}), W \subseteq R_{2}[x]$

$p_{1}(x)= 4x^2+3x^3$

$p_{2}(x)= 1+2x^2+3x^3$

$p_{3}(x) = 3-2x^2+3x^3$

From $p_{1},p_{2},p_{3}$ choose a basis $B$ for $W$ and state why $B$ is a basis for $W$.

And what does this $R_{2}[x]$ mean?

I really hope you can give a detailled answer, I will also reward that answer with a bounty because I need to know how to solve tasks like that!

cnmesr
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    It is given that $W$ is spanned by $p_1,p_2,p_3$, all you got to do is check if they are linear independent. $R_2[x]$ is the vector space of all polynomial of degree $\leq 2$ – Itay4 Mar 13 '17 at 20:09
  • @Itay4 But how can I check that? I cannot form these to vectors / matrix, or can I? Why is the polynomial of degree $\leq 2$ but we have degree $3$ there? – cnmesr Mar 13 '17 at 20:14
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    Check this http://math.stackexchange.com/a/1350678/385242 – Itay4 Mar 13 '17 at 20:22
  • As for the degree, this is the notation I know, your's probably different. I guess it's of degree $\leq 3$, even though it doesn't make much sense. – Itay4 Mar 13 '17 at 20:30

1 Answers1

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The symbol $\mathbb{R}_2[x]$ usually denotes the space of polynomials (in the variable $x$ with coefficients in $\mathbb{R}$) of degree $\leq 2$. Since in your case, some of your polynomials are of degree three, I'll assume that this is a typo in the original question and that the intent was to write $\mathbb{R}_3[x]$.

Assuming that, you are given three vectors $p_1,p_2,p_3$ in some vector space $\mathbb{R}_3[x]$ and consider the vector space $W$ spanned by them. This is a vector space of dimension $3$ or less. To determine the precise dimension of $W$ and a basis, we need to understand whether the vectors $(p_1,p_2,p_3)$ are linearly independent and if not, choose a subset which is linearly independent.

In your case, $(p_1)$ alone is linearly independent because it is non-zero. Next, $(p_1,p_2)$ is a linearly independent list because $p_2$ is not a scalar multiple of $p_1$ ($p_1$ has no free coefficient while $p_2$ has a free coefficient). As for $(p_1,p_2,p_3)$, we can see that

$$ 3 - 2x^2 + 3x^3 = 3(1 + 2x^2 + 3x^3) - 2(4x^2 + 3x^3) $$

so $p_3 = 3p_1 - 2p_2$ is a linear relation showing that $p_3$ is dependent on $(p_1,p_2)$. Hence, $W$ is two dimensional with a basis given by $(p_1,p_2)$.

levap
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  • Thank you very much for your answer :) You call $p_1,p_2,p_3$ vectors. Does that mean I can write them like this too: $p_{1}=\begin{pmatrix} 0\ 4x^2\ 3x^3 \end{pmatrix}, p_2=\begin{pmatrix} 1\ 2x^2\ 3x^3 \end{pmatrix}, p_3= \begin{pmatrix} 3\ -2x^2\ 3x^3 \end{pmatrix}$? Can I even leave the $x$ variable away? – cnmesr Mar 13 '17 at 21:41
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    @cnmesr: No. A vector is just an element of some vector space, it doesn't have to look like an array. What you can do is identify the vectors in $\mathbb{R}_3[x]$ with columns vectors in $\mathbb{R}^4$ so that $a + bx + cx^2 + dx^3$ is identified with $\begin{pmatrix} a \ b \ c \ d \end{pmatrix}$ and solve the corresponding problem there. – levap Mar 13 '17 at 21:46
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    So it would be possible (maybe not really helpful in this example) to solve it like that as well? Write them as vectors as you have shown above this comment and check for linear (in)dependency? $u\begin{pmatrix} 0\ 0\ 4\ 3 \end{pmatrix}+v\begin{pmatrix} 1\ 0\ 2\ 3 \end{pmatrix}w\begin{pmatrix} 3\ 0\ -2\ 3 \end{pmatrix}=\begin{pmatrix} 0\ 0\ 0\ 0 \end{pmatrix}$ Thank you a lot by the way, I will give you the bounty when it's available :) – cnmesr Mar 13 '17 at 22:18
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    @cnmesr: Yes, it is indeed possible and sometimes useful. You can always translate any question in linear algebra in some (maybe strange) vector space to a question in the "standard" vector space $\mathbb{F}^n$. – levap Mar 13 '17 at 22:35