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In a vector space $P_3[\mathbb R]$ vectors $p_1(x)=1,p_2(x)=3x^2,p_3(x)=x+x^2-3x^3$ are given. Determine a vector $p_4(x)$ such that vectors $p_1,p_2,p_3,p_4$ are linearly independent.

If we choose a standard basis for $P_3[\mathbb R]$, $B=\{1,x,x^2,x^3\}$, we can represent vectors in explicit form (with components): $p_1(x)=(1,0,0,0),p_2(x)=(0,0,3,0),p_3(x)=(0,1,1,-3),p_4(x)=(u,v,w,z)$

$$\alpha(1,0,0,0)+\beta(0,0,3,0)+\gamma(0,1,1,-3)+\delta(u,v,w,z)=(0,0,0,0)$$

Is it possible to find the coordinates of vector $p_4$ from this equation?

If not, then what method to use?

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  • $\begingroup$ You can go like that but it will get a bit messy. For me the easyer way is to put those vectors in a determinant and find for which u,v,w,z is the determinant non-zero. $\endgroup$ – CTSnake Apr 24 '17 at 19:48
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Your approach will not really work. To say that the set is linearly independent means that your equation must imply $\alpha=\beta=\gamma=\delta=0$. So you aren't solving the equation; you are trying to find values for some of the variables such that there is only one solution for all the other variables. There is no obvious way to do this. In fact, there are infinitely many different possible values of $p_4$ that would work, so you cannot expect to solve this in any deterministic way.

I would suggest instead just trying to guess a value and then check that it works. The set $\{p_1,p_2,p_3,p_4\}$ will be linearly independent as long as $p_4$ is not a linear combination of $p_1$, $p_2$, and $p_3$, so most values of $p_4$ will work. Just pick some $p_4$ that does not appear to be a linear combination of the others, and then see if you can prove that it really does give a linearly independent set.

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  • $\begingroup$ CTSnake suggested a method with determinant. Is that method sufficient (does it work)? Then we won't have to guess anything. $\endgroup$ – user300046 Apr 24 '17 at 20:56
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    $\begingroup$ That is one way to test whether a value of $p_4$ works. You still won't be able to use it to "solve" for $p_4$ though, since infinitely many different values work: instead you might guess a value and then use the determinant to test whether it works. $\endgroup$ – Eric Wofsey Apr 24 '17 at 20:57
  • $\begingroup$ From a determinant method, $z\neq 3v\Rightarrow z\neq 0,v\neq 0$. If we set $z=1,v=1$ and $u=w=0$, then one solution for $p_4$ is $p_4=(0,1,0,1)$. $\endgroup$ – user300046 Apr 24 '17 at 21:11
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Assuming that this isn’t a trick question so that the original three vectors are linearly independent to begin with (easily verified) a simple way to go is to take a vector in the orthogonal complement of the span of the three coefficient vectors that you’ve got. That’s guaranteed to be linearly independent of the others. Since the null space of a matrix is the orthogonal complement of its row space, write the three vectors as the rows of a matrix and compute the rref: $$\begin{bmatrix}1&0&0&0\\0&0&3&0\\0&1&1&-3\end{bmatrix}\to\begin{bmatrix}1&0&0&0\\0&1&0&-3\\0&0&1&0\end{bmatrix}$$ from which you can read that $(0,3,0,1)^T$, i.e. $3x+x^3$ will do the trick.

If you want to find all vectors that are linearly independent of the first three, then the determinant method suggested by CTSnake is a good way to go.

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