Given a vector ${\bf x}\in V$ that has components $\xi_1,\xi_2,\ldots,\xi_n$ with respect to a basis ${\bf e_1,e_2,\ldots,e_n}$, construct a new basis such that the components of $\bf x$ with respect to this basis is 1,0,0,...,0

Maybe I am missing something obvious here but any basis that looks like $$ {\bf x,e_1,e_2,...,e_{i-1},e_{i+1},...,e_n} $$ would be a valid answer right?

For instance

$$ {\bf x} = (1){\bf x} + (0){\bf e_1} + (0){\bf e_2} + (0){\bf e_4} +(0) {\bf e_5} + \ldots + (0){\bf e_n} $$

Clearly ${\bf x},{\bf e_1},{\bf e_2},\ldots,{\bf e_{i-1}},{\bf e_{i+1}},\ldots,{\bf e_n}$ is a linearly independent set of vectors that span the space ${\bf V}$ right?

What am I doing wrong?

  • 2
    $\begingroup$ As long as $x\neq 0$ and you pick $i$ such that $\xi_i \neq 0$, this is fine. Good work! $\endgroup$ – Cameron Williams Oct 1 '15 at 4:00
  • $\begingroup$ @CameronWilliams Is it possible to consider the set $\{x\}$ and then extend this to get a basis for $V.$ Then we would have a basis in which the components of $x$ are $(1,0,...0).$ Does this make sense? $\endgroup$ – model_checker Jan 11 '18 at 16:45
  • 1
    $\begingroup$ @SuperMario Absolutely. You could extend it by looking for all of the vectors that are orthogonal to $x$. $\endgroup$ – Cameron Williams Jan 11 '18 at 17:51

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