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Let $E=(e_1, e_2,e_3,e_4)$ be the standard basis in $\mathbb R^4$ and let another basis be given by $$ B = (\begin{pmatrix} 1\\ 0\\ 0\\ 0 \end{pmatrix} ,\begin{pmatrix} 2\\ 1\\ 0\\ 0 \end{pmatrix} ,\begin{pmatrix} 0\\ 3\\ 1\\ 0 \end{pmatrix} ,\begin{pmatrix} 0\\ 0\\ 4\\ 1 \end{pmatrix} ).$$

What are the change of basis matrices from E to B and from B to E? Also why? Thanks for the help!

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  • $\begingroup$ What you wrote is no basis of $\,\Bbb R^4\,$ but simply a matrix, so: what did you exactly mean? To take the matrix's columns as vectors...? $\endgroup$ – DonAntonio Feb 10 '13 at 20:47
  • $\begingroup$ yes, the columns are the basis vectors, sorry $\endgroup$ – Thomas Feb 10 '13 at 20:51
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Hint: Given an arbitrary vector ($\alpha_1, \ldots)$ in basis $E$, you want to find the representation of that same vector in basis $B$: $$\sum_i \alpha_i e_i = \sum_i \beta_i b_i.$$ What does this equation look like in matrix notation? How would you solve for the unknown $\beta_i$ if I give you $\alpha_i$?

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  • $\begingroup$ This was on a test, and I said that the change of basis matrix of the change from E to B is given by $B^{-1}$ and the change of basis matrix from B to E is given by B. I got zero points. $\endgroup$ – Thomas Feb 10 '13 at 21:17
  • $\begingroup$ You should go talk to the professor/TA; unless you are leaving out some parts of the problem (ie, maybe you were suppose to explicitly work out $B^{-1}$), that should be correct. $\endgroup$ – user7530 Feb 10 '13 at 21:34

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