I guys, I'm having a hard time working on the following exercise:


$$L_A:\mathbb{R}^2\rightarrow\mathbb{R}^2,\;\mathcal x\mapsto A\cdot\mathcal x$$

$\begin{array}{l}with\;respect\;to\;the\;stan dard\;basis\;\begin{pmatrix}1\\0\end{pmatrix},\begin{pmatrix}0\\1\end{pmatrix}\;for\;\mathbb{R}^2.\;Find\;the\;transformed\;matrix\;T\;such\;that\;the\;same\;map\;L_A\;can\;be\;expressed\;as\\\\\end{array}$

$$L_A:\mathbb{R}^2\rightarrow\mathbb{R}^2,\;\mathcal x\mapsto T\cdot\mathcal x$$

$\begin{array}{l}with\;respect\;to\;the\;non-stan dard\;basis\;\begin{pmatrix}1\\1\end{pmatrix},\begin{pmatrix}0\\1\end{pmatrix}\;for\;\mathbb{R}^2.\\\end{array}$

So far, I have I have multiplied the matrix A times each basis, and then I tried to find a matrix T, s.t. T times the non-standard basis is equal to the first linear map. I am not sure if that's the good way to do it.

Can someone help me? Thank you!

EDIT: I have found the matrix $T=\begin{pmatrix}0&1\\-1&3\end{pmatrix}$. With this T if I multiply by the non-standard basis, the result is the same as in the first linear map.

  • 1
    $\begingroup$ take a look here math.stackexchange.com/questions/2560162/… $\endgroup$ – user Jan 24 '18 at 16:32
  • $\begingroup$ @gimusi I have edited my post. $\endgroup$ – TTT Jan 24 '18 at 16:42
  • $\begingroup$ The matrix $T$ that you’ve come up with says that the basis vector $(1,1)^T$ is mapped to $-1$ times the other basis vector, i.e., $(0,-1)^T$ in the standard basis. Is that what $A(1,1)^T$ gives you? $\endgroup$ – amd Jan 24 '18 at 19:00
  • $\begingroup$ @amd when I multiply the matrix T times the non-standard basis, I get the same result as the matrix A times the standard basis. Is my answer correct? the matrix T is what should be done? I don't think I have understood well the question. $\endgroup$ – TTT Jan 24 '18 at 23:51
  • $\begingroup$ The columns of a transformation matrix are the images of the basis vectors. For whatever ordered basis $(v_1,v_2)$ the matrix $T$ might refer to, its first column says that the transformation represented by $T$ maps $v_1$ to $-v_2$. You have $v_1=(1,1)^T$ and $v_2=(0,1)^T$ w/r to the standard basis, so if $T$ is correct, then it must be the case that $A(1,1)^T=-(0,1)$. It does not, so your matrix $T$ is wrong. $\endgroup$ – amd Jan 25 '18 at 0:46

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