I have been trying the solve the following question without any success:

Let $\lambda_1, \lambda_2, \lambda_3$ be three distinct complex numbers and define the polynomials $m(\lambda), m_1(\lambda), m_2(\lambda), m_3(\lambda), l_1(\lambda), l_2(\lambda), l_3(\lambda)$ as follows:

$m(\lambda) = (\lambda - \lambda_1)(\lambda - \lambda_2)(\lambda - \lambda_3)$

$m_1(\lambda) = (\lambda - \lambda_2)(\lambda - \lambda_3)$

$m_2(\lambda) = (\lambda - \lambda_1)(\lambda - \lambda_3)$

$m_3(\lambda) = (\lambda - \lambda_1)(\lambda - \lambda_2)$

$l_1(\lambda) = \frac{m_1(\lambda)}{m_1(\lambda_1)}$

$l_2(\lambda) = \frac{m_2(\lambda)}{m_2(\lambda_2)}$

$l_3(\lambda) = \frac{m_3(\lambda)}{m_3(\lambda_3)}$

a) Show that $l_i(\lambda_j) = 1$ if $i = j$ and 0 if $i \neq j$. This is easy, just plug in the values and we get the answer.

b) Using the above result show that $l_1 , l_2 , l_3$ are linearly independent in the vector space $\mathcal{C_2}[\lambda]$, of all polynomials in $\lambda$ of degree less than or equal to 2.

For (b) part, I am running into complicated equations in showing that if $\alpha*l_1 + \beta*l_2 + \gamma*l_3 = 0$, then all $\alpha, \beta, \gamma$ are zero.

I am sure there must be an elegant way of solving this. Can anyone point me in the right direction?



Assume $al_1+bl_2+cl_3=0$, where $0$ is the zero function. Then it is true that $$al_1(\lambda)+bl_2(\lambda)+cl_3(\lambda)=0$$ for every $\lambda$. Conclude by evaluating in $\lambda_i$ for $i=1,2,3$.

  • $\begingroup$ Oh.. thanks so much, I missed that the equation should be true for every lambda $\endgroup$ – user94300 Sep 2 '15 at 7:14

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