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[Attempted solution[1]I attempted this question but just wanted to check if I'm on the right track.

I started the question by :

Let { ($a_o$+$a_1$$x^2$), ($a_o$-$a_1$$x^2$), ($a_1$$x^2$-$a_0$), ($a_1$$x^2$-$a_2$x) ∈ $P_2$

Followed by putting in vector form and into a matrix, reducing it to row echelon form and comparing the rank to the number of columns to determine it is lenearly dependent. Am I right to do it this way?

Jenny
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  • What dimension is $P_2$? How many linearly independent vectors can be in an $n$ dimensional space? – operatorerror Apr 22 '18 at 04:43
  • would the dimension be 4? – Jenny Apr 22 '18 at 04:44
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    To begin with, $P_2$ is the set of all polynomials with real coefficients of degree at most $2$, i.e. $P_2={a_0+a_1x+a_2x^2~:~a_1,a_2,a_3\in\Bbb R}$. It looks like you are maybe looking at a very specific set of four polynomials, or maybe you are intending to take four arbitrary polynomials but you are missing a term from each... in either event this would be incorrect. Note that $P_2$ is of dimension ___ because of _____ and so if you have four polynomials since four is greater than _____ this implies that ______ because of ______. – JMoravitz Apr 22 '18 at 04:45
  • @Jenny not quite. How did you arrive at that number? – operatorerror Apr 22 '18 at 04:45
  • I got the number from the question, since it said any four polynomials so i thought it would be 4 sets and thus the dimension would be 4 – Jenny Apr 22 '18 at 04:49
  • @Jenny In a space of dimension 4, you could have 4 linearly independent vectors, but not 5. That's what the dimension means, essentially. – saulspatz Apr 22 '18 at 04:51
  • The dimension of a vector space is the number of vectors in a basis for the space. The vectors in a basis for a space $V$ must be linearly independent and must span the entirety of $V$. For example, the basis ${[1,0],[0,1]}$ acts as a basis for $\Bbb R^2$ so $\Bbb R^2$ is of dimension $2$. A common beginner exercise is to prove that any two bases for the same space must have the same number of vectors, so defining dimension in this way is unambiguous. Another common exercise is to show that any set of $n$ vectors in an $m$-dimensional space where $n>m$ must be linearly dependent. – JMoravitz Apr 22 '18 at 04:53
  • I attached an image of my attempted solution. Would this be the right way to address the question? – Jenny Apr 22 '18 at 06:33
  • No. I will reiterate and reemphasize my earlier point. You have shown that a particular and very specific collection of four vectors in $P_2$ happen to be linearly dependent. This is not sufficient proof to show that all collections of four vectors in $P_2$ are linearly dependent. Proof by example works only to show that there exists at least one situation where the claim is true, but it does not work for addressing the claim that it is true always for all situations. – JMoravitz Apr 22 '18 at 14:50
  • Something else weird and awkward about your attempt is that you end with the line "$rank(A)=3\neq \text{# of columns}\implies\text{linearly dependent}$... but $3$ is the number of columns... with the way you set it up you mean to say that it is not the number of rows. Keep track of the difference between setting your vectors to test independence as the rows versus as the columns and what the result of the row reduction will give you. If you have trouble remembering, think of one of the easier examples of using the vector $[1,1,1,1]$. – JMoravitz Apr 22 '18 at 14:55

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