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Find a maximal linearly independent set in the following subset $C$ of $$\Bbb R^{k+1}:C= \{ (1,t,t^2,t^3,\ldots,t^k): t \in \Bbb R\}~,$$ where $k$ is an integer and $k \ge$ 1.

Please tell me if my approach below is correct.

Consider the $n+1$ vectors obtained by putting $t = 1,2,3, \ldots ,k+1$. Now we construct a matrix with these vectors as follows:

$$\begin{bmatrix} 1&1&1&1&1& \cdots &\cdots&1\\ 1&2&2^2&2^3&2^4& \cdots & \cdots &2^k\\ 1&3&3^2& \cdots & \cdots & \cdots & &3^k\\ \vdots & \vdots &&&&& \ddots& \vdots\\ 1&(k+1)&(k+1)^2& \cdots & \cdots & \cdots & \cdots &(k+1)^k\end{bmatrix}_{(k+1)\text{by}(k+1)}$$

Now we consider the columns of this matrix. My claim that the columns are linearly independent which implies the vectors i have chosen are linearly independent.

Denote the columns as $a_1, a_2,\ldots, a_{k+1}$. Let there be $k+1$ real scalars $\beta_1,\beta_2, \ldots,\beta_{k+1}$ such that $$\sum_{i=1}^{k+1} a_i\beta_i = \theta$$

Then the new vector on LHS is $$\begin{aligned} \bigl( &\beta_1 + \beta_2 \cdot 1 +\beta_3 \cdot 1^2 + \dotsb + \beta_{k+1} \cdot 1^k, \\ &\beta_1 + \beta_2 \cdot 2 +\beta_3 \cdot 2^2 + \dotsb + \beta_{k+1} \cdot 2^k, \\ &\hspace{36pt}\vdots \\ &\beta_1 + \beta_2 \cdot (k+1)+ \dotsb + \beta_{k+1} \cdot (k+1)^k \bigr) \end{aligned}$$ We see that each component of this vector is equal to $0$ which tells me that the polynomial of degree $k$
$$\beta_1 + \beta_2 \cdot t +\beta_3 \cdot t^2 + \dotsb + \beta_{k+1} \cdot t^k$$ has $k+1$ distinct zeros.

This means that the polynomial must be zero hence all the scalars must be zero and hence the column vectors are linearly independent.

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  • $\begingroup$ Maximal LI subset is basis of C $\endgroup$ – Tojrah Apr 25 at 18:04
  • $\begingroup$ Where did you find this problem? What have you tried and where did you get stuck? $\endgroup$ – Servaes Apr 25 at 18:07
  • $\begingroup$ you may want to check Vandermonde matrices and their properties (en.wikipedia.org/wiki/Vandermonde_matrix) $\endgroup$ – P. Quinton Apr 25 at 19:19
  • $\begingroup$ Yes vandermonde matrix makes the proof easier . $\endgroup$ – Souvik Deb Apr 26 at 18:27

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