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Given the basis $\beta = \{ (1, 1, 0),\ (1, 0, -1),\ (2, 1, 0)\}$ and the matrix:

$$ A = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $$

with respect to $\beta$.

How to change this matrix with respect to standard basis?

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The change matrix $P$ from the standard basis $\alpha$ to the basis $\beta$ is $$P=\left(\begin{matrix}\\1&1&2\\ 1&0&1\\ 0&-1&0 \end{matrix}\right)$$ hence the linear transformation which's represented by the matrix $A$ in the basis $\beta$ is represented in the standard basis by the matrix $$PAP^{-1}$$

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  • $\begingroup$ Wait. Look answers in math.stackexchange.com/questions/340978/change-of-basis-matrix. If you say that P changes a column of standard coordinates $v_{stanard}$ to a column of $\beta$-coordinates, $v_\beta = P v_{standard}$ and question says that A applies to $\beta$-coordinates, then application should look like $A v_\beta = A P v_{standard}$. We have AP or $P^{-1}AP$, if inverse conversion to standard is prefixed. I do not understand why you get $PAP^{-1}$ from. $\endgroup$ – Val Jun 3 '14 at 20:29
  • $\begingroup$ @Val I know that the appellation makes some confusion but this is what's used by the most books: the change matrix $P$ from a basis $\alpha$ to a basis $\beta$ is a matrix that changes a column of $\beta-$ coordinates to a column of $\alpha-$ coordinates i.e. $v_{\alpha}=Pv_{\beta}$. $\endgroup$ – user63181 Jun 3 '14 at 20:46
  • $\begingroup$ @user63181 Ok, I see. They say $\vec v = E[\vec v]_e = EPP^{-1}[\vec v]_e = B [\vec v]_b$ so that basis is translated by right-multiplying with P, $B = EP$, and coordinates are translated covarianly, $[\vec v]_b = P^{-1} [\vec v]_e$. $\endgroup$ – Val Jun 4 '14 at 11:57

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