# 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
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 $$\displaystyle{T}_{{{2}}}\ :\ {V}\rightarrow{V}.$$

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 $$\displaystyle{\left(-{a}{v}_{{{1}}}-{9}{a}{v}_{{{2}}}\right)}$$
Hence, the linear transformation $$\displaystyle{\left({T}_{{{2}}}{T}_{{{1}}}\right)}{\left({v}\right)}$$ is $$\displaystyle{\left(-{a}{v}_{{{1}}}-{9}{a}{v}_{{{2}}}\right)}$$

### Relevant Questions

Guided Proof Let $${v_{1}, v_{2}, .... V_{n}}$$ be a basis for a vector space V.
Prove that if a linear transformation $$T : V \rightarrow V$$ satisfies
$$T (v_{i}) = 0\ for\ i = 1, 2,..., n,$$ then T is the zero transformation.
To prove that T is the zero transformation, you need to show that $$T(v) = 0$$ for every vector v in V.
(i) Let v be the arbitrary vector in V such that $$v = c_{1} v_{1} + c_{2} v_{2} +\cdots + c_{n} V_{n}$$
(ii) Use the definition and properties of linear transformations to rewrite T(v) as a linear combination of $$T(v_{j})$$ .
(iii) Use the fact that $$T (v_{j}) = 0$$
to conclude that $$T (v) = 0,$$ making T the zero transformation.

Let $$\left\{v_{1},\ v_{2}, \dots,\ v_{n}\right\}$$ be a basis for a vector space V. Prove that if a linear transformation $$\displaystyle{T}\ :\ {V}\rightarrow\ {V}$$ satisfies $$\displaystyle{T}{\left({v}_{{{1}}}\right)}={0}\ \text{for}\ {i}={1},\ {2},\dot{{s}},\ {n},$$ then T is the zero transformation.
Getting Started: To prove that T is the zero transformation, you need to show that $$\displaystyle{T}{\left({v}\right)}={0}$$ for every vector v in V.
(i) Let v be an arbitrary vector in V such that $$\displaystyle{v}={c}_{{{1}}}\ {v}_{{{1}}}\ +\ {c}_{{{2}}}\ {v}_{{{2}}}\ +\ \dot{{s}}\ +\ {c}_{{{n}}}\ {v}_{{{n}}}.$$
(ii) Use the definition and properties of linear transformations to rewrite $$\displaystyle{T}\ {\left({v}\right)}$$ as a linear combination of $$\displaystyle{T}\ {\left({v}_{{{1}}}\right)}$$.
(iii) Use the fact that $$\displaystyle{T}\ {\left({v}_{{i}}\right)}={0}$$ to conclude that $$\displaystyle{T}\ {\left({v}\right)}={0}$$, making T the zero tranformation.

The dominant form of drag experienced by vehicles (bikes, cars,planes, etc.) at operating speeds is called form drag. Itincreases quadratically with velocity (essentially because theamount of air you run into increase with v and so does the amount of force you must exert on each small volume of air). Thus
$$\displaystyle{F}_{{{d}{r}{u}{g}}}={C}_{{d}}{A}{v}^{{2}}$$
where A is the cross-sectional area of the vehicle and $$\displaystyle{C}_{{d}}$$ is called the coefficient of drag.
Part A:
Consider a vehicle moving with constant velocity $$\displaystyle\vec{{{v}}}$$. Find the power dissipated by form drag.
Express your answer in terms of $$\displaystyle{C}_{{d}},{A},$$ and speed v.
Part B:
A certain car has an engine that provides a maximum power $$\displaystyle{P}_{{0}}$$. Suppose that the maximum speed of thee car, $$\displaystyle{v}_{{0}}$$, is limited by a drag force proportional to the square of the speed (as in the previous part). The car engine is now modified, so that the new power $$\displaystyle{P}_{{1}}$$ is 10 percent greater than the original power ($$\displaystyle{P}_{{1}}={110}\%{P}_{{0}}$$).
Assume the following:
The top speed is limited by air drag.
The magnitude of the force of air drag at these speeds is proportional to the square of the speed.
By what percentage, $$\displaystyle{\frac{{{v}_{{1}}-{v}_{{0}}}}{{{v}_{{0}}}}}$$, is the top speed of the car increased?
Express the percent increase in top speed numerically to two significant figures.

Consider $$\displaystyle{V}={s}{p}{a}{n}{\left\lbrace{\cos{{\left({x}\right)}}},{\sin{{\left({x}\right)}}}\right\rbrace}$$ a subspace of the vector space of continuous functions and a linear transformation $$\displaystyle{T}:{V}\rightarrow{V}$$ where $$\displaystyle{T}{\left({f}\right)}={f{{\left({0}\right)}}}\times{\cos{{\left({x}\right)}}}−{f{{\left(π{2}\right)}}}\times{\sin{{\left({x}\right)}}}.$$ Find the matrix of T with respect to the basis $$\displaystyle{\left\lbrace{\cos{{\left({x}\right)}}}+{\sin{{\left({x}\right)}}},{\cos{{\left({x}\right)}}}−{\sin{{\left({x}\right)}}}\right\rbrace}$$ and determine if T is an isomorphism.

To determine:
To prove:
The function $$\displaystyle{T}:{V}\rightarrow{W}$$ is linear transformation if and only if
$$\displaystyle{T}{\left({a}{u}\ +\ {b}{v}\right)}={a}{T}{\left({u}\right)}\ +\ {b}{T}{\left({v}\right)}$$ for all vectors u and v and all scalars a and b.
Two hollow metal spheres are concentric with each other. Theinner sphere has a radius of 0.131 m and a potential of 89.6 V. Theradius of the outer sphere is 0.155 m and its potential is 78.3 V.If the region between the spheres is filled with Teflon, find theelectric energy contained in this space.

For the V vector space contains all $$\displaystyle{2}\times{2}$$ matrices. Determine whether the $$\displaystyle{T}:{V}\rightarrow{R}^{{1}}$$ is the linear transformation over the $$\displaystyle{T}{\left({A}\right)}={a}\ +\ {2}{b}\ -\ {c}\ +\ {d},$$ where $$A=\begin{bmatrix}a & b \\c & d \end{bmatrix}$$

All bases considered in these are assumed to be ordered bases. In Exercise, compute the coordinate vector of v with respect to the giving basis S for V. V is $$R^2, S = \left\{ \begin{bmatrix}1 \\ 0 \end{bmatrix}\begin{bmatrix} 0 \\1 \end{bmatrix} \right\}, v = \begin{bmatrix} 3 \\-2 \end{bmatrix}$$

To show:
The set $$\displaystyle{\left\lbrace{T}{\left({x}_{{1}}\right)},\ \ldots\ ,{T}{\left({x}_{{k}}\right)}\right\rbrace}$$ is a linearly independent subset of $$\displaystyle{R}^{{{m}}}$$
Given:
Let $$\displaystyle{T}\ :\ {T}\ :\ {R}^{{{n}}}\rightarrow{R}^{{{m}}}$$ be a linear transformation with nulity zero. If $$\displaystyle{S}={\left\lbrace{x}_{{{1}}},\ \cdots\ \ ,{x}_{{{k}}}\right\rbrace}$$ is a linearly independent subset of $$\displaystyle{R}^{{{n}}}.$$
(a) What is the acceleration when the car is at B located at an angle of 35.0°? Express your answer in terms of the unit vectors $$\displaystyle\hat{{{i}}}$$ and $$\displaystyle\hat{{{j}}}$$.
1. (Enter in box 1) $$\displaystyle\frac{{m}}{{s}^{{2}}}\hat{{{i}}}+{\left({E}{n}{t}{e}{r}\in{b}\otimes{2}\right)}{P}{S}{K}\frac{{m}}{{s}^{{2}}}\hat{{{j}}}$$
4. ( Enter in box 4) $$\displaystyle\frac{{m}}{{s}^{{2}}}\hat{{{i}}}+$$
5. ( Enter in box 5) $$\displaystyle\frac{{m}}{{s}^{{2}}}\hat{{{j}}}$$