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
Best Answer
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 $4$.