Answer to Question #238059 in Linear Algebra for okuhle

Question #238059

1.   The figure shows two bases, consisting of unit vectors, for ll?2: S —— (i,j) and 0 = (ut,ut).


(a)    Find the transition matrix  s B

(b)    Find the transition matrix PB S



1
Expert's answer
2021-11-11T11:32:11-0500


We can see that the question is not clear as there are some typing errors.

so we can manipulate it and provide a idea how to solve these types of questions>


Adjusting the conditions of the questions, and rewriting it as;


The figure show two basis for R2:S=(i,j)\reals^2:S=(i,j) and B=(u1,u2)B=(u_1',u_2')


1) Find the transition matrix PSBP_S\to\>_B

2) Find the transition matrix PBSP_B\to\>_S







Now we can see our Solution:


In reference to the figure, first basis S=(i,j)S=(i,j)


Where i=[10]i=\begin{bmatrix} 1 \\ 0 \end{bmatrix} ,j=[01]\quad\>j=\begin{bmatrix} 0 \\ 1 \end{bmatrix}


Second basis B(u1,u2)B(u'_1,u'_2)


Where u1=(120)u'_1=\begin{pmatrix} \frac{1}{2} \\ 0 \end{pmatrix} , u2=(12)u'_2= \begin{pmatrix} -1 \\ 2 \end{pmatrix}



a) Expressing elements of BB in terms of elements of SS



(120)\begin{pmatrix} \frac{1}{2} \\ 0 \end{pmatrix} =a(10)=a\begin{pmatrix} 1 \\ 0 \end{pmatrix} +b(01)+b\begin{pmatrix} 0 \\ 1 \end{pmatrix}


    a=12,b=0\implies a=\frac{1}{2},b=0



(12)\begin{pmatrix} -1 \\ 2 \end{pmatrix} =a(10)=a'\begin{pmatrix} 1 \\ 0 \end{pmatrix} +b(01)+b'\begin{pmatrix} 0\\ 1 \end{pmatrix}


     a=1,b=2\implies\>a'=-1,b'=2



\therefore Transition matrix is (aabb)\begin{pmatrix} a & a' \\ b& b' \end{pmatrix}


\therefore Transition matrix PSB=(12102)P_S\>\to_B=\begin{pmatrix} \frac{1}{2} & -1\\ 0& 2 \end{pmatrix}


Thus we found PS matrix.



b) Transition matrix PBSP_B\>\to\>S =(PSB)1=(P_S\to\>_B)^{-1}


(12102)\begin{pmatrix} \frac{1}{2} & -1 \\ 0& 2 \end{pmatrix} 1^{-1} =11(21012)= \frac{1}{1}\begin{pmatrix} 2 & 1 \\ 0 & \frac{1}{2} \end{pmatrix}


PBS\therefore P_B\to\>_S =(21012)= \begin{pmatrix} 2 & 1 \\ 0 & \frac{1}{2} \end{pmatrix}


thus we found PB Matrix.


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