I understand how we can find the transformation matrix $D$ with respect to another basis $B$, by using a transformation matrix that we already know, say $A$: $$D = C^{-1}\cdot A\cdot C$$

Where $C$ is the change of basis for the basis $B$ and $C^{-1}$ is its inverse.

Now, this is fine, but there's a concrete problem that gives different results if we first multiply $A.C$ rather than $C^{-1} \cdot A$

I am watching a video, and specifically at min 10.25, the professor starts by first multiplying $A.C$ and not $C^{-1}.A$, which is what I did, obtaining at the end a different $D$.

My questions are:

  • How the order that we use to multiply the matrices can change the resulting $D$ transformation matrix?

  • Which version is correct:

    1. Multiply first $C^{-1}$ by $A$ or multiply first $A$ by $C$, and why?

I have used Maxima to solve this problem and apparently it has done what I did: it has multiplied first $C^{-1}$ by $A$...obtaining at the end the same matrix I obtained, as you can see from the picture:

enter image description here

Just for completeness:

$$B = \{ \left[ \begin{matrix} 1 \\ 2 \end{matrix}\right], \left[ \begin{matrix} 2 \\ 1 \end{matrix}\right] \}$$

$$A = \left[ \begin{matrix} 3 & -2 \\ 2 & -3 \end{matrix}\right]$$

  • 1
    $\begingroup$ Do you mean $C^{-1}(AC)\ne (C^{-1}A)C$? That is impossible. According to associative law, they are equal. $\endgroup$ – KittyL Jun 3 '15 at 13:55
  • 2
    $\begingroup$ In your picture, your $A$ is apparently different from the $A$ in the video. $\endgroup$ – KittyL Jun 3 '15 at 13:56
  • 1
    $\begingroup$ @KittyL Oh my god, you are right, the second entry of the second row is $-2$ and not $-3$... $\endgroup$ – nbro Jun 3 '15 at 13:58
  • $\begingroup$ @KittyL Now, maxima gives the same $D$ matrix as in the video. Thanks! Lol, just for one number, I had to write a 10 min markdown/latex question... $\endgroup$ – nbro Jun 3 '15 at 13:59
  • $\begingroup$ For some reason it's very easy to copy down wrong numbers when doing linear algebra. So check that first. $\endgroup$ – KittyL Jun 3 '15 at 14:03

Your Answer

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

Browse other questions tagged or ask your own question.