Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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...? – DonAntonio Feb 10 '13 at 20:47
yes, the columns are the basis vectors, sorry – Thomas Feb 10 '13 at 20:51

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 $\alpha_i$?

share|cite|improve this answer
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. – Thomas Feb 10 '13 at 21:17
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. – user7530 Feb 10 '13 at 21:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.