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Let $V$ be a vector space and let $\{ \alpha_1 ,\alpha_2, ..., \alpha_n \}$ be a basis of $V$. For all $i=1, 2,...,n;$ we define $\beta_i=\alpha_i+...+\alpha_n$.

a) Is $\{\beta_1, \beta_2, ...,\beta_n\}$ a basis of $V$ (Prove the answer)

b) Which is the matrx of basis change?

Can someone help me with this excercise, i have no idea were to begin

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    $\begingroup$ do you know the definitions? start there! $\endgroup$ – Ittay Weiss Jun 17 '14 at 5:19
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Try to first get some intuition by specializing the problem. Take a vector space you know well, like $\mathbb R^3$, and choose a simple basis you know and love, for instance the standard basis. Now, construct the $\beta _i$ (there are only three to construct) and check if it's a basis or not. That way you will gain a better understanding of the question and you might start getting ideas on how to prove your answer.

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Consider the transformation $T:V\to V$ that sends $v_i\to v_i+\cdots+v_n$. Then its matrix in base $B=\{v_1,\ldots,v_n\}$ is $$\begin{pmatrix}1&0&\cdots&0&0\\1&1&\cdots&0&0\\\vdots&\vdots&\ddots&\vdots&\vdots\\1&1&\cdots&1&0\\1&1&\cdots&1&1\end{pmatrix}$$

What is its determinant?

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For question $(a$ , consider any linear combination $$\Sigma_{i=1,2,..,n} c_i\beta_i $$ of the $\beta_i$ , you can see that, if you substitute back the values $\beta_i=\alpha_i+...+\alpha_n$ in the linear combination above, this will give you a linear combination $$\Sigma_{i=1,2.,,n}c_i(\Sigma \alpha_i) $$

of the original basis vectors. This linear combination can only equal zero if all the $c_i's $ are zero, because...

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For (a) from user99680's answer $\beta _i$s won't be a basis iff $\sum \beta _i c_i =0$ and $c_i \ne 0$ for some $i$.

$$0 = \sum c_i\beta_i = \sum ^nc_i \sum _{j=i} ^n \alpha_j = \sum _{i=1}^n \alpha_i\sum_{j=1}^ic_i = \sum\alpha_id_i $$, where, $d_i = \sum _{j=1}^ic_j$ which is false unless $d_1=d_2=...=d_n=0 \implies c_1 = c_2 = c_3 = .... = 0$

For b) Pedro Tamaroff has already answered (I think its slightly off), so I'll just correct it here:

$$ T=\begin{pmatrix}1&1&\cdots&1&1\\0&1&\cdots&1&1\\\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\cdots&1&1\\0&0&\cdots&0&1\end{pmatrix} $$

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  • $\begingroup$ That is also my answer for a). $\endgroup$ – user99680 Jun 17 '14 at 6:12
  • $\begingroup$ Oh. Sorry. Missed it. Will edit my answer. $\endgroup$ – tpb261 Jun 17 '14 at 6:14
  • $\begingroup$ No problem; thanks. $\endgroup$ – user99680 Jun 17 '14 at 6:18

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