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Consider $T:\mathbb{R^3}\rightarrow \mathbb{R^3}$ a linear transformation with matricial representation (in the canonical basis of $\mathbb{R^3}$:

$A = \begin{bmatrix} 1 & 0 & 0 \\[0.3em] 0 & 1 & 2 \\[0.3em] 0 & 0 & 1 \end{bmatrix}$

Find the matrix that represents $T$ in the respect to the basis $b=<(1,1,1), (1,1,0), (1,0,0)$

so i'm not understanding how to do this...

I tried to calculate the changing basis matrix (from the canonical basis to the basis we want). But now what to I do with that matrix? I thought in might be useful but the fact that we are trying to calculate the transformation in another basis is making me confused :/

By the way the matrix of changing basis that i reached to was:

$B = \begin{bmatrix} 1 & 1 & 1 \\[0.3em] 1 & 1 & 0 \\[0.3em] 1 & 0 & 0 \end{bmatrix}$

Can someone clarify to me how should I proceed?

Thank you very much!

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  • $\begingroup$ Maybe you can try to write down the image of a random vector in $\mathbb{R^3}$, say $x=(x_1,x_2,x_3)$. Then ask yourself how you can write this vector in the new basis, call this one $y$. And solve the linear system of equations $Ax=y$. $\endgroup$ Commented May 23, 2016 at 16:59
  • $\begingroup$ How is that going to help me @Tim Huijgens? I'm sorry I'm really feeling blank with this I need a more clear explanation... I'm not getting there alone $\endgroup$ Commented May 23, 2016 at 17:05
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    $\begingroup$ The idea is that in an other basis we want a matrix $C$ which represents the same transformation as the matrix $A$ represents in the old basis. This can only be the case if $C$ maps all vectors in $\mathbb{R^3}$ to the same images as $A$. Therefore write down the image of $x$, this represents all images from $A$. Then write this down in the new basis-form, with new coördinates, calling it $y$. Finally we solve $Cx=y$ and we get $B$. If all works out correctly your answer should be $BAB^{-1}$. en.wikipedia.org/wiki/Change_of_basis $\endgroup$ Commented May 23, 2016 at 17:15
  • $\begingroup$ Are you working with column vectors, in which case you left-multiply by the matrix, or with row vectors, in which case you right-multiply? It makes a difference, and unfortunately I can’t tell from $B$ since that’s symmetric. $\endgroup$ Commented May 23, 2016 at 19:43

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Your matrix should look like $$T=BA(B)^{-1}$$

It's the general formula.

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    $\begingroup$ This answer is incorrect if we’re using the convection of left-multiplying by the matrix to apply the transformation. $A(1,1,1)^T=(1,3,1)^T$, which is $(1,2,-2)^T$ relative to $b$. The first column of $BAB^-1$, however, is something else entirely. $\endgroup$ Commented May 23, 2016 at 19:35
  • $\begingroup$ If the vectors are $(e_i)_j$, isn't the most general linear transformation $(e'_k)_l=P_{klij}(e_i)_j$ ? $\endgroup$ Commented Oct 26, 2022 at 5:41

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