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In basis $\langle e_1,e_2,e_3,e_4 \rangle$ linear transformation $\varphi$ has matrix:

$$\begin{pmatrix}1&2&0&1\\3&0&-1&2\\2&5&3&1\\1&2&1&3\end{pmatrix}$$

find matrix of this transformation in basis $\langle e_1,e_1+e_2,e_1+e_2+e_3,e_1+e_2+e_3+e_4 \rangle$

What steps should I reproduce? I am a bit new to this

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Sorry but I disagree with PJK because the vectors $\varphi(e_1)$, $\varphi(e_1+e_2)$, $\varphi(e_1+e_2+e_3)$, and $\varphi(e_1+e_2+e_3+e_4)$ have to be expressed in the new basis $\langle b_1,b_2,b_3,b_4\rangle$ with $b_1= e_1,b_2= e_1+e_2, b_3= e_1+e_2+e_3$ and $b_4 = e_1+e_2+e_3+e_4$.

This process of chancing the basis can be made like this:

  1. Call $P$ the matrix of the new basis in the old one ($b_i$ expressed in columns relatively to the $e_i$)
  2. Call $A$ the matrix of your question.
  3. The matrix of $\varphi$ in the new basis $\langle b_1,b_2,b_3,b_4\rangle$ is: $$ B = P^{-1}AP $$

The calculus are for you! ;-)

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  • $\begingroup$ Of course you are right. I forgot about changing the coordinates. Thanks for pointing it out. $\endgroup$ – PJK Jun 14 '17 at 16:13
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We see from the matrix you give us that: φ(e_1)=1*e_1+3*e_2+2*e_3+1*e_4 φ(e_2)=2*e_1+0*e_2+5*e_3+2*e_4 φ(e_3)=0*e_1+(-1)*e_2+3*e_3+1*e_4 φ(e_4)=1*e_1+2*e_2+1*e_3+3*e_4

because φ is linear fuction φ(e_1+e_2)=φ(e_1)+φ(e_2) and φ(e_1+e_2+e_3)=φ(e_1)+φ(e_2)+φ(e_3) and φ(e_1+e_2+e_3+e_4)=φ(e_1)+φ(e_2)+φ(e_3)+φ(e_4)

then you have to calculate the: φ(e_1),φ(e_2),φ(e_1+e_2),φ(e_1+e_2+e_3),φ(e_1+e_2+e_3+e_4) and then write the as linear combination of e_1,e_2,e_1+e_2,e_1+e_2+e_3,e_1+e_2+e_3+e_4 and write the coefficients as column as i wrote you above with the other base

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