I have a linear algebra question from a practice sheet that I'm completely lost on. New to linear algebra, so any guidance in how to get started or what to look for would be really appreciated. The question's as follows:

Let $K$ be a field and let $V$ be the subspace of $K[x]$ consisting of all polynomials of degree at most 4. Let $p_1{(x)},p_2{(x)}, ... , p_5{(x)}$ be distinct polynomials in $V$ that satisfy $p_i{(0_K)} = 1_K$ for each $1 \leq i \leq 5$. Is the set {$p_1(x), ... , p_5(x)$} necessarily linearly dependent?

Again thanks in advance for any help.

  • 4
    $\begingroup$ What about $p_1(x)=1$ and $p_i(x)=1+x^{i-1}$ for $2\le i\le 5$ ? $\endgroup$ – Adren Jan 22 '17 at 21:21
  • 2
    $\begingroup$ This Question poses a contrast with the case that five polynomials $p_i$ all satisfy $p_i(1_K) = 0_K$. This homogeneous constraint would force a linear dependence relation. $\endgroup$ – hardmath Jan 22 '17 at 21:39

Let's consider a few things.

First, what is the maximum length of a list which could remain linearly independent in $K[X]$. Obviously we can say that $(1, t, t^2, t^3, t^4)$ spans $K[X]$, which means that it is possible to have a list of length 5 which is linearly independent in consideration. (Note that if the spanning set had had 4 or fewer elements you would have been able to conclude immediately that linear dependence was necessary). With that check out of the way, it is not immediately clear what to check next; as such, you may wish to try and construct a counter-example.

Is it possible to construct $p_1(x), ... , p_5(x)$ such that they are linearly independent. If so, you can conclude that no, it is not necessary for them to be linearly dependent. If you fail to construct such an example, it may illustrate along the way a manner for us to prove that they must be linearly dependent.

If we start from the spanning set above $1, t, t^2, t^3, t^4$ we may be able to work to get these to fit, without changing the linear dependence. We can take $p_1(x) = 1$ which of course satisfies the necessary $p_1(0) = 1$. However, taking $p_2(x) = x^2$ is not sufficient. Instead, we would need to add $1$ to $p_2$ as defined above in order to have it belong to $V$. We could use the same process to consider $p_3(x)...p_5(x)$.

Now we have the following:

$p_1(x) = 1$

$p_2(x) = t + 1$

$p_3(x) = t^2 + 1$

$p_4(x) = t^3 + 1$

$p_5(x) = t^4 + 1$

Is that set of polynomials linearly independent? There are a number of ways to check for this, I will include a less-than elegant check to demonstrate. I would urge you to consider other alternatives for this consideration.

Take $a_1...a_5$ to be the coefficients such that, no matter the value of $x$, we have $a_1*p_1(x) + ... + a_5*p_5(x) = 0$.

\begin{align} a_1*p_1(x) + a_2*p_2(x) + a_3*p_3(x) + a_4*p_4(x) + a_5*p_5(x) &= 0 \\ a_1 + a_2*(x + 1) + a_3*(x^2 + 1) + a_4*(x^3 + 1) + a_5*(x^4 + 1) &= 0 \\ a_2x + a_3x^2 + a_4x^3 + a_5x^4 + (a_1 + a_2 + a_3 + a_4 + a_5) &= 0 \\ \\ \textbf{Take x to be 1} \\ a_2(1) + a_3(1)^2 + a_4(1)^3 + a_5(1)^4 + a_1 + a_2 + a_3 + a_4 + a_5 &= 0\\ a_1 + 2a_2 + 2a_3 + 2a_4 + 2a_5 &= 0 \\ a_1 &= -2(a_2 + a_3 + a_4 + a_5) \\ \\ \textbf{Take x to be 0} \\ a_2(0) + a_3(0)^2 + a_4(0)^3 + a_5(0)^4 + a_1 + a_2 + a_3 + a_4 + a_5 &= 0\\ a_1 + a_2 + a_3 + a_4 + a_5 &= 0 \\ a_1 &= -(a_2 + a_3 + a_4 + a_5) \\ \\ y &:= a_2 + a_3 + a_4 + a_5 \\ -2y &= -y \\ 2y &= y \\ y &= 0 \implies a_1 = 0 \end{align}

Now, not only do we know that $a_1 = 0$ but also that $a_2 + a_3 + a_4 + a_5 = 0$. So we could rewrite the relationship as $a_2x + a_3x^2 + a_4x^3 + a_5x^4 = 0$. Recognizing that this set is linearly independent is enough to conclude that the above set is as well.

I again do not recommend you using this methodology to test for linear independence. If you have worked with determinants that is a more straightforward manner of demonstrating that this set is linearly independent. I have shown it to you in this manner to hopefully convince you of the answer, but leave room for you to work through the problem in a more elegant manner.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.