To find: The linear transformation (T_2T_1)(v) for an arbitrary vector v in V. The vectors {v_1,v_2} is vasis for the vector space V. Given: The linear transformation with satisfying equations T_{1}(v_1)=3v_{1}+v_{2}, T_{1}(v_1)=-3v_{1}+v_{2}, T_{2}(v_{1})=-5v_{2}, and T_{2}(v_2)=-v_{1}+6v_{2} are given as T_{1} : V rightarrow V and T_{2} : V rightarrow V.

To find: The linear transformation (T_2T_1)(v) for an arbitrary vector v in V. The vectors {v_1,v_2} is vasis for the vector space V. Given: The linear transformation with satisfying equations T_{1}(v_1)=3v_{1}+v_{2}, T_{1}(v_1)=-3v_{1}+v_{2}, T_{2}(v_{1})=-5v_{2}, and T_{2}(v_2)=-v_{1}+6v_{2} are given as T_{1} : V rightarrow V and T_{2} : V rightarrow V.

Question
Transformation properties
asked 2021-02-09
To find:
The linear transformation \(\displaystyle{\left({T}_{{2}}{T}_{{1}}\right)}{\left({v}\right)}\) for an arbitrary vector v in V.
The vectors \(\displaystyle{\left\lbrace{v}_{{1}},{v}_{{2}}\right\rbrace}\) is vasis for the vector space V.
Given:
The linear transformation with satisfying equations \(\displaystyle{T}_{{{1}}}{\left({v}_{{1}}\right)}={3}{v}_{{{1}}}+{v}_{{{2}}},\)
\(\displaystyle{T}_{{{1}}}{\left({v}_{{1}}\right)}=-{3}{v}_{{{1}}}+{v}_{{{2}}},\ {T}_{{{2}}}{\left({v}_{{{1}}}\right)}=-{5}{v}_{{{2}}},\) and \(\displaystyle{T}_{{{2}}}{\left({v}_{{2}}\right)}=-{v}_{{{1}}}+{6}{v}_{{{2}}}\) are given as
\(\displaystyle{T}_{{{1}}}\ :\ {V}\ \rightarrow{V}\) and \(\displaystyle{T}_{{{2}}}\ :\ {V}\rightarrow{V}.\)

Answers (1)

2021-02-10
Calculation:
To solve for \(\displaystyle{T}_{{{2}}}{\left({T}_{{1}}{\left({v}\right)}\right)}\) find \(\displaystyle{T}_{{{1}}}{\left({v}\right)}\) use properties of linear transformation.
\(\displaystyle{T}_{{{1}}}{\left({v}\right)}={T}_{{{1}}}{\left({a}{v}_{{{1}}}+{b}{v}_{{{2}}}\right)}\)
\(\displaystyle{T}_{{{1}}}{\left({v}\right)}={a}{T}_{{{1}}}{\left({v}_{{{1}}}\right)}+{b}{T}_{{{1}}}{\left({v}_{{{2}}}\right)}\)
Substitute \(\displaystyle{3}{v}_{{{1}}}+{v}_{{{2}}}\) for \(\displaystyle{T}_{{{1}}}{\left({v}_{{{1}}}\right)}\) and 0 for \(\displaystyle{T}_{{{1}}}{\left({v}_{{{2}}}\right)}\) in the above expression.
\(\displaystyle{T}_{{{1}}}{\left({v}\right)}={a}{\left({3}{v}_{{{1}}}+{v}_{{{2}}}\right)}+{b}{\left({0}\right)}\)
\(\displaystyle={a}{\left({3}{v}_{{{1}}}+{v}_{{{2}}}\right)}\)
Substitute \(\displaystyle{a}{\left({3}{v}_{{{1}}}+{v}_{{{2}}}\right)}\) for \(\displaystyle{T}_{{{1}}}{\left({v}\right)}\) to find \(\displaystyle{T}_{{{2}}}{\left({T}_{{{1}}}{\left({v}\right)}\right)}\) and it is written as,
\(\displaystyle{T}_{{{2}}}{\left({T}_{{{1}}}{\left({v}\right)}\right)}={T}_{{{2}}}{\left({a}{\left({3}{v}_{{{1}}}+{v}_{{{2}}}\right)}\right)}\)
\(\displaystyle{T}_{{{2}}}{\left({T}_{{{1}}}{\left({v}\right)}\right)}={3}{a}{T}_{{{2}}}{\left({v}_{{{1}}}\right)}+{a}{T}_{{{2}}}{\left({v}_{{{2}}}\right)}\)
Substitute \(\displaystyle-{5}{v}_{{{2}}}\) for \(\displaystyle{T}_{{{2}}}{\left({v}_{{{1}}}\right)}\) and \(\displaystyle-{v}_{{{1}}}+{6}{v}_{{{2}}}\) for \(\displaystyle{T}_{{{2}}}{\left({v}_{{{2}}}\right)}\) in the above expression.
\(\displaystyle{T}_{{{2}}}{\left({T}_{{{1}}}{\left({v}\right)}\right)}={3}{a}{\left(-{5}{v}_{{{2}}}\right)}+{a}{\left(-{v}_{{{1}}}+{6}{v}_{{{2}}}\right)}\)
\(\displaystyle-{15}{a}{v}_{{{2}}}-{a}{v}_{{{1}}}+{6}{a}{v}_{{{2}}}\)
\(\displaystyle=-{a}{v}_{{{1}}}\ -\ {9}{a}{v}_{{{2}}}\)
Therefore, the linear transformation \(\displaystyle{\left({T}_{{{2}}}{T}_{{{1}}}\right)}{\left({v}\right)}\) is \(\displaystyle{\left(-{a}{v}_{{{1}}}-{9}{a}{v}_{{{2}}}\right)}\)
Answer:
Hence, the linear transformation \(\displaystyle{\left({T}_{{{2}}}{T}_{{{1}}}\right)}{\left({v}\right)}\) is \(\displaystyle{\left(-{a}{v}_{{{1}}}-{9}{a}{v}_{{{2}}}\right)}\)
0

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