1
$\begingroup$

Let $A= \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ be a transformation $T:\mathbb{R}^3\to \mathbb{R}^3$ by the elementary basis $B$ and let $V=\{(1 ,0 ,0),(1 ,1 ,0),(0 ,0 ,1)\}$ basis such that the $A$ is diagonal.

I am trying to show that $[I]^{B}_{V}[T]_{B}=[T]_V$

So we have to solve

$(1,0,0)=\alpha(1 ,0 ,0)+\beta(1 ,1 ,0)+\gamma(0 ,0 ,1)$

$(0,1,0)=\alpha(1 ,0 ,0)+\beta(1 ,1 ,0)+\gamma(0 ,0 ,1)$

$(0,01,1)=\alpha(1 ,0 ,0)+\beta(1 ,1 ,0)+\gamma(0 ,0 ,1)$

which is $\begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ but $\begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}\cdot \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}\neq \text{diagonal}$

$\endgroup$
  • $\begingroup$ For a base change you have to compute $SAS^{-1}$ and not $A^2$. $\endgroup$ – Dietrich Burde May 28 '18 at 15:14
  • $\begingroup$ So $S=[I]^B_V$ and $S^{-1}=[I]_{B}^V$? $\endgroup$ – newhere May 28 '18 at 15:17
  • $\begingroup$ See for example here, how it works. $\endgroup$ – Dietrich Burde May 28 '18 at 15:19
  • $\begingroup$ Yes, exactly. $\, $ $\endgroup$ – Berci May 28 '18 at 15:20
1
$\begingroup$

Let

$$S= \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

then by the change of basis $y=Sv$ and $x=Su$, we have that

$$y=Ax \implies Sv=ASu \implies v=S^{-1}ASu$$

$\endgroup$

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.