Calculation:
The function is defined as,
\(\displaystyle{T}{\left({x},{y}\right)}={\left(\sqrt{{{x}}},{x}{y},\sqrt{{{y}}}\right)}\)
Assume two general vectors \(\displaystyle{u}={\left({u}_{{{1}}},{u}_{{{2}}}\right)}\) and \(\displaystyle{v}={\left({v}_{{{1}}},{v}_{{{2}}}\right)}\)
Then \(\displaystyle{u}+{v}={\left({u}_{{{1}}}+{v}_{{{1}}},{u}_{{{2}}}+{v}_{{{2}}}\right)}\)
\(\displaystyle{c}{u}={\left({c}{u}_{{{1}}},{c}{u}_{{{2}}}\right)}\)
The function is a linear transformation if it satisfies the two properties as mentioned in the approach part.
Compute \(\displaystyle{T}{\left({u}+{v}\right)}\) and \(\displaystyle{T}{\left({u}\right)}+{T}{\left({v}\right)}\) as,
\(\displaystyle{T}{\left({u}+{v}\right)}={T}{\left({u}_{{{1}}}+{v}_{{{1}}},{u}_{{{2}}}+{v}_{{{2}}}\right)}\)
\(\displaystyle={\left(\sqrt{{{u}_{{{1}}}+{v}_{{{1}}}}},{\left({u}_{{{1}}}+{v}_{{{1}}}\right)}{\left({u}_{{{2}}}+{v}_{{{2}}}\right)},\sqrt{{{u}_{{{2}}}+{v}_{{{2}}}}}\right)}\)
\(\displaystyle={\left(\sqrt{{{u}_{{{1}}}+{v}_{{{1}}}}},{\left({u}_{{{1}}}{u}_{{{2}}}+{u}_{{{1}}}{y}_{{{2}}}+{u}_{{{2}}}{v}_{{{1}}}+{v}_{{{1}}}{v}_{{{2}}}\right)},{\left(\sqrt{{{u}_{{{2}}}+{v}_{{{2}}}}}\right)}\right.}\)
\(\displaystyle{T}{\left({u}\right)}+{T}{\left({v}\right)}={T}{\left({u}_{{{1}}},{u}_{{{2}}}\right)}+{T}{\left({v}_{{{1}}},{v}_{{{2}}}\right)}\)
PSK\(=(\sqrt{u_{1}},u_{1}u_{2},\sqrt{u_{2}})+(\sqrt{v_{1}},v_{1}v_{2},\sqrt{v_{2}})\)
\(\displaystyle={\left(\sqrt{{{u}_{{{1}}}}}+\sqrt{{{v}_{{{1}}}}},{u}_{{{1}}}{u}_{{{2}}}+{v}_{{{1}}}{v}_{{{2}}},\sqrt{{{u}_{{{2}}}}}+\sqrt{{{v}_{{{2}}}}}\right.}\)
Since \(\displaystyle{T}{\left({u}+{v}\right)}\ne{q}{T}{\left({u}\right)}+{T}{\left({v}\right)}\), the first property is not satisfied.
Therefore, the function is not a linear transformation.
Whether the function is a linear transformation or not. T : R^{2} rightarrow R^{3}, T(x,y)=(sqrt{x},xy,sqrt{y})
Whether the function is a linear transformation or not. T : R^{2} rightarrow R^{3}, T(x,y)=(sqrt{x},xy,sqrt{y})
Answers (1)
Relevant Questions
\(\displaystyle{T}\ :\ {R}^{{{2}}}\rightarrow{R}^{{{2}}},{T}{\left({x},{y}\right)}={\left({x},{y}^{{{2}}}\right)}\)
The transformation \(T\ :\ \mathbb{R}^{2}\ \rightarrow\ \mathbb{R}^{2}\) sending the origin to itself and a triangle of vertices
\((0,\ 0),\ (1,\ 0),\ (0,\ 1)\) to a triangle of verices (0, 0),
\((\sqrt{\frac{2}{2}},\ \sqrt{\frac{2}{2}}),\ (- \sqrt{\frac{2}{2}},\ \sqrt{2}).\)
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 \(\displaystyle{R}^{{{n}}}.\)
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.
Consider the linear transformation \(\displaystyle{U}:{R}^{{3}}\rightarrow{R}^{{3}}\) defined by \(U \left(\begin{array}{c}x\\ y \\z \end{array}\right) = \left(\begin{array}{c} z - y \\ z + y \\ 3z - x - y \end{array}\right)\) and the bases \(\epsilon = \left\{ \left(\begin{array}{c}1\\ 0 \\0\end{array}\right), \left(\begin{array}{c}0\\ 1 \\ 0\end{array}\right), \left(\begin{array}{c}0\\ 0 \\ 1\end{array}\right) \right\}, \gamma = \left\{ \left(\begin{array}{c}1 - i\\ 1 + i \\ 1 \end{array}\right), \left(\begin{array}{c} -1\\ 1 \\ 0\end{array}\right), \left(\begin{array}{c}0\\ 0 \\ 1\end{array}\right) \right\}\), Compute the four coordinate matrices \(\displaystyle{{\left[{U}\right]}_{{\epsilon}}^{{\gamma}}},{{\left[{U}\right]}_{{\gamma}}^{{\gamma}}},\)
Prove whether \({f}:\mathbb{R}\to\mathbb{R}\ \text{defined by}\ f{{\left({x}\right)}}={2}{x}\) is a linear transformation.
Which one equivalent to the linear transformation \({T}:\mathbb{R}\to\mathbb{R}\ \text{defined by}\ {T}{\left({1}\right)}={2}?\)
Consider the following linear transformation \(T : P_2 \rightarrow P_3\), given by \(T(f) = 3x^2 f\)'. That is, take the first derivative and then multiply by \(3x^2\) (a) Find the matrix for T with respect to the standard bases of \(P_n\): that is, find \([T]_{\epsilon}^{\epsilon}\), where- \(\epsilon = {1, x, x^2 , x^n}\) (b) Find N(T) and R(T). You can either work with polynomials or with their coordinate vectors with respect to the standard basis. Write the answers as spans of polynomials. (c) Find the the matrix for T with respect to the alternate bases: \([T]_A^B\) where \(A = {x - 1, x, x^2 + 1}, B = {x^3, x, x^2, 1}.\)
