# To find maximal linearly independent subset of an infinite subset of vectors

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.

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