Another homework problem is asking me the following:
Given basis B = {[2 -3 0], [0 1 1], [4 0 -1]} and standard basis E = {[1 0 0], [0 1 0], [0 0 1]}, find the change of coordinates matrix M^B B sub E. Then show how you would use this to help calculate the basis B coordinates of vector [1 3 -2].
So far, I have used the formula in Theorem 15 for this basis to try and set [x]_E equal to P_E<-B, but I am missing [x].
An example I found online helped me to realize I need to use the basis B in this, but then it said something about multiplication with another matrix that wasn't an identity matrix (E in this problem). Don't I need to have an original matrix for this? If not, am I supposed to multiply basis B by the identity?
Any pointers would be greatly appreciated.
*Side note, I'm not sure if it matters, but I was unable to get the vectors to look like a column vector as they do in the problem on the handout I received.
Best Answer
Getting the hang of changing basis can be really tricky so no worries about asking this question at all! Might I recommend watching the Youtuber "3blue1brown" and looking at his playlist essence of algebra. This video gives a really nice explanation for what is going on when we change basis: https://www.youtube.com/watch?v=P2LTAUO1TdA
I will just give you answers for your problem with a little explanation:
Firstly the matrix B that has columns equal to your basis vectors is : $$ B = \begin{pmatrix} 2 & 0 & 4 \\ -3 & 1 & 0 \\ 0 & 1 & -1 \\ \end{pmatrix} $$ Let us think about what B, as a map/function/transformation does. It takes some vector $v_B$ ( a vector writen in our B basis ) and gives us the vector $v_E$ (the vector written in the standard basis B) Try it, pick any vector, something simple like $(1,0,0)$ or $(1,1,0)$ and notice that $B(v_B) := Bv_B = v_E$ So "B takes us from our weird basis B and gives us a vector in the standard basis"
However we are after something a little different indeed. We wish to travel from the standard basis into the the world of basis B. We want to go the other way around than what the matrix $\textbf{B} $ does. So how can we do this? Why invert the matrix of course! A little algebra gives:
$$ B^{-1} = \frac{1}{14}\begin{pmatrix} 1 & -4 & 4 \\ 3 & 2 & 12 \\ 3 & 2 & -2 \\ \end{pmatrix} $$
And hence $B^{-1}(v_E) := B^{-1}v_E = v_B $.
$B^{-1}$ takes a vector in the standard basis and gives us the same vector but in basis B So to find the basis B coordinates of vector $(1, 3, -2)^T$ We do $B^{-1} (1, 3, -2)^T $ =
$$ \frac{1}{14}\begin{pmatrix} 1 & -4 & 4 \\ 3 & 2 & 12 \\ 3 & 2 & -2 \\ \end{pmatrix} \begin{pmatrix} 1 \\ 3 \\ -2 \\ \end{pmatrix} = \frac{1}{14} \begin{pmatrix} -19 \\ -15 \\ 13 \\ \end{pmatrix} = v_B $$
A simple check of our answer for $v_B $ shows it yields the correct result!
Hope this helped. Please if you have an issues with my soloution post them below! Good luck with studies
Oskar