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 "\\reals^2:S=(i,j)" and "B=(u_1',u_2')"


1) Find the transition matrix "P_S\\to\\>_B"

2) Find the transition matrix "P_B\\to\\>_S"







Now we can see our Solution:


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


Where "i=\\begin{bmatrix}\n 1 \\\\\n 0 \n\\end{bmatrix}" ,"\\quad\\>j=\\begin{bmatrix}\n 0 \\\\\n 1\n\\end{bmatrix}"


Second basis "B(u'_1,u'_2)"


Where "u'_1=\\begin{pmatrix}\n \\frac{1}{2} \\\\\n 0\n\\end{pmatrix}" , "u'_2= \\begin{pmatrix}\n -1 \\\\\n 2 \n\\end{pmatrix}"



a) Expressing elements of "B" in terms of elements of "S"



"\\begin{pmatrix}\n \\frac{1}{2} \\\\\n 0\n\\end{pmatrix}" "=a\\begin{pmatrix}\n 1 \\\\\n 0 \n\\end{pmatrix}" "+b\\begin{pmatrix}\n 0 \\\\\n 1 \n\\end{pmatrix}"


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



"\\begin{pmatrix}\n -1 \\\\\n 2\n\\end{pmatrix}" "=a'\\begin{pmatrix}\n 1 \\\\\n 0\n\\end{pmatrix}" "+b'\\begin{pmatrix}\n 0\\\\\n 1 \n\\end{pmatrix}"


"\\implies\\>a'=-1,b'=2"



"\\therefore" Transition matrix is "\\begin{pmatrix}\n a & a' \\\\\n b& b'\n\\end{pmatrix}"


"\\therefore" Transition matrix "P_S\\>\\to_B=\\begin{pmatrix}\n \\frac{1}{2} & -1\\\\\n 0& 2\n\\end{pmatrix}"


Thus we found PS matrix.



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


"\\begin{pmatrix}\n \\frac{1}{2} & -1 \\\\\n 0& 2\n\\end{pmatrix}" "^{-1}" "= \\frac{1}{1}\\begin{pmatrix}\n 2 & 1 \\\\\n 0 & \\frac{1}{2}\n\\end{pmatrix}"


"\\therefore P_B\\to\\>_S" "= \\begin{pmatrix}\n 2 & 1 \\\\\n 0 & \\frac{1}{2}\n\\end{pmatrix}"


thus we found PB Matrix.


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