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We are given the following matrix: $$ A=\begin{pmatrix} b_1 & \lambda_1b_1 & \lambda_1^2b_1 & \ldots &\lambda_1^{n-1}b_1 \\ b_2 & \lambda_2b_2 & \lambda_2^2b_2 & \ldots &\lambda_2^{n-1}b_2\\ \vdots\\ b_n & \lambda_nb_n & \lambda_n^2b_n & \ldots &\lambda_n^{n-1}b_n \\ \end{pmatrix}. $$

For what values of $\lambda_k$ and $b_k\,(k=1,\ldots,n)$ is the rank of $A$ equal to $n$?

How does one solve this and what is the solution?

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  • $\begingroup$ See here for how to type matrices. $\endgroup$
    – user1551
    Dec 2 '12 at 22:24
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$\det(A)=b_1\cdots b_n \det(V)$, where $\det(V)$ is the determinant of the Vandermonde matrix of $\lambda_1,\cdots,\lambda_n$, so $\det(A)=b_1\cdots b_n \prod\limits_{1\le i< j \le n}(\lambda_j - \lambda_i)$, so if $b_i \neq 0$ and $\lambda_j \neq \lambda_i$ ($i \neq j$), then $A$ is full rank.

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  • $\begingroup$ Should it be $A = b_1...b_2det(V)$ or b_n? $\endgroup$
    – Deeya
    Dec 2 '12 at 23:12
  • $\begingroup$ @Deeya it should be $b_n$, edited it. $\endgroup$
    – chaohuang
    Dec 2 '12 at 23:15
  • $\begingroup$ Great, thank you so much. :) $\endgroup$
    – Deeya
    Dec 2 '12 at 23:17

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