3
$\begingroup$

Let $p_1=t^5+t^4,p_2=t^5-7t^3, p_3=t^5-1,p_4=t^5+3t \in\mathbb Q_{\le5}[t]$. Show $(p_1,p_2,p_3,p_4)$ is linearly independent.

My attempt with the definition of linearly independence: Let $\lambda_{1,2,3,4}\in \mathbb Q$ and consider $0=\lambda_1(t^5+t^4)+\lambda_2(t^5-7t^3)+\lambda_3(t^5-1)+\lambda_4(t^5+3t)$ By plugging in $t=0$ we get $\lambda_3=0$. Then we remain with $0=(\lambda_1+\lambda_2+\lambda_4)t^5+\lambda_1t^4-7\lambda_2t^3-3\lambda_4t$, but I do not see how to continue here. I tried to plug in some more values for $t$.

First question: Is this attempt in general ok and how could one continue?

Another idea is to consider a basis of $\mathbb Q_{\le5}[t]$ i.e. $\cal B=\{1,t,t^2,t^3,t^4,t^5\}$. Then the transformation matrix is given by

$ \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 3 \\ 0&0&0&0 \\ 0 & -7 & 0 & 0 \\ 1 &0 &0 &0 \\ 1 &1 &1 &1 \end{pmatrix} $

Its rank is $4$.

Second question: Since its rank is 4 can we conclude that $(p_1,p_2,p_3,p_4)$ is linear independent?

Third question: Are there other possibilites to check the linear independence here?

$\endgroup$
1
$\begingroup$

First, you can divide your equality with $t$, thus obtaining:

$$0 = (\lambda_1 + \lambda_2 + \lambda_4)t^4 + \lambda_1t^3 - 7\lambda_2t^2 - 3\lambda_4$$

Plugging in $t = -1$ gives

$$0 = -6\lambda_2-2\lambda_4 \implies \lambda_4 = -3\lambda_2$$

Plugging in $t = 1$ gives

$$0 = 2\lambda_1 -6\lambda_2 - 2\lambda_4 = 2\lambda_1 \implies \lambda_1= 0$$

Finally, plugging in $t = 2$ gives

$$0 = -12\lambda_2 + 13\lambda_4$$

which together with $\lambda_4 = -3\lambda_2$ gives $\lambda_2 = \lambda_4 = 0$.

Your second method is also correct, and indeed it yields that the given polynomials are linearly independent.

$\endgroup$
0
$\begingroup$

Each given polynomial in the form $a+bt+ct^2+dt^3+et^4+ft^5$ in the basis $\cal B=\{1,t,t^2,t^3,t^4,t^5\}$ can be represented by its components (a,b,c,d,e,f).

Consider the matrix of the components (it is not a transormation matrix) and reduce to RREF.

If we obtain rank=4 then yes the polynomials are linearly independent.

The other approach is to write the linear combination and show that it is zero if and only if the coefficients are all zero. This approach is completely equivalent to the previous since rank=4 implies that $Ax=0 \iff x=0$.

$\endgroup$
2
  • $\begingroup$ ok yes that is what I meant. I wrote its matrix of components. Its rank is clearly 4. So we are done $\endgroup$ – user519338 Mar 19 '18 at 20:21
  • 1
    $\begingroup$ @J.D I didn't check the derivation but if the calculation is correct yes you are done! $\endgroup$ – user Mar 19 '18 at 20:22
0
$\begingroup$

In your first approach you must remember that the $0$ at LHS in $$0=\lambda_1(t^5+t^4)+\lambda_2(t^5-7t^3)+\lambda_3(t^5-1)+\lambda_4(t^5+3t)$$ is the zero polynomial, that is the polinomial that is null for all values of $t$. So , reordering the RHS we find: $$0=t^5(\lambda_1+\lambda_2+\lambda_3+\lambda_4 )+\lambda_1t^4-7\lambda_2t^3+3\lambda_4t -\lambda_3$$ and this is the null plynomial only if $\lambda_1=\lambda_2=\lambda_3=\lambda_4=0$

$\endgroup$

Your Answer

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