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So the one below is not a matrix they are vectors. Im still not used to writing vectors.

So Im trying to to find the values of k for which the vectors are linearly independent. Well what I know is Ax = 0 means linear independence.

The thing is. I've never handled polynomials and its really messing with me. Yes this is homework and I will be referencing this site and the contributors.

$$ \begin{matrix} 1 & k & 2k+1 & 3+k & 4/(1+k^4) \\ 2 & k^2 & 4k^2 -2 & 2k+5 & (5/(3+k^2)) \\ 3 & k^3 & 8k^8+3k & k+3 & 6/(5+2k^2) \\ 4 & k^4 & 16k^{16} - 4k^3 -1 & -4k^2 & 7/(1+k^2) \\ \end{matrix} $$

Show $$ ({4x^3, 3x^2, 2x, 1}) \\ spans \ p^3 \ (vector\ space\ of\ polynomials\ of \ degree\ at\ most\ 3) $$

does $$(8.9x^3+31.95x^3, 0.5x^3+x^2-0.6x+7.77, - 100x+55)$$ also span all of $$P^3$$

If you guys can give me links about materials regarding this topic I would greatly appreciate it as I cannot seem to find polynomials inside a vector in my linear algebra book. Thank you

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No, $Ax=0$ does not mean linear independance. Linear independance of the column vectors is: $AX=0 \;\Rightarrow\; X=0$. –  1015 Jan 30 '13 at 20:14

1 Answer 1

1) You have $5$ column vectors in $\mathbb{R}^4$ whose dimension is $4$. So they must be linearly dependant, for every $k$.

2) Yes, $(4x^3,3x^2,2x,1)$ span $P_3$, since you can write every polynomial of dergree $3$ or less: $$ a_3x^3+a_2x^2+a_1x+a_0=\frac{a_3}{4}4x^3+\frac{a_2}{3}3x^2+\frac{a_1}{2}2x+a_0\cdot 1. $$

3) For your other set of polynomials, note that there are only three of them. So their span has dimension $3$ or less. Therefore it can not be equal to $P_3$ whose dimension is $4$.

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Ok each vector is written vertically. Each vector is a 4x1 vector. So youre saying if I want it to be linearly independent it always has to be a square matrix? (1,2,3,4) , (k,K^2,K^3,K^4), (2k+1, 4k^2-2, 8k^8+3k, 16k^(16) - 4k^3-1) , (3+k, 2k+5, k+3, -4k^2), (4/(1+4k), 5/(3+k^2), 6/(5+2k^2), 7/(1+k^2)) –  user1730308 Jan 30 '13 at 21:08
    
Ok Lets say I remove two vectors and Im Left with the last three vector. This is not a square matrix but it can be part of Rn right? So how would you find k then in this case? Do you just put this in an augmented matrix and figure out how to get rid of the extra powers of k? –  user1730308 Jan 30 '13 at 21:15

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