0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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 $Ax=y$. $\endgroup$ – Tim Huijgens May 23 '16 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 $\endgroup$ – Granger Obliviate May 23 '16 at 17:05
  • 1
    $\begingroup$ 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 $\endgroup$ – Tim Huijgens May 23 '16 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. $\endgroup$ – amd May 23 '16 at 19:43
0
$\begingroup$

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

It's the general formula.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – amd May 23 '16 at 19:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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