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Problem list Free problems

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Problem: Find out if the transformations are linear: \[ \begin{array}{l} A x=\left(3 x_{1}-2 x_{2}-x_{3}, 1, x_{1}+2 x_{2}+3\right), \\ B x=\left(3 x_{1}-2 x_{2}-x_{3}, 1, x_{1}^{3}+2 x_{2}+3 x_{3}\right), \\ C x=\left(3 x_{1}-2 x_{2}-x_{3}, x_{3}, x_{1}+2 x_{2}+3 x_{3}\right) . \end{array} \]

1.7.1 Linear transformations

1.28 $

Problem: Given the coordinates of the vector \( x=\{2,5,10\} \) in the basis \( \left(e_{1}, e_{2}, e_{3}\right) \). Find its coordinates in the basis \( \left(e_{1}{ }^{\prime}, e_{2}{ }^{\prime}, e_{3}{ }^{\prime}\right) \), where \[ \left\{\begin{array}{c} e_{1}{ }^{\prime}=e_{1}+e_{2}+6 e_{3} \\ e_{2}{ }^{\prime}=\frac{6}{5} e_{1}-e_{2} \\ e_{3}{ }^{\prime}=-e_{1}+e_{2}+e_{3} \end{array} .\right. \]

1.7.3 Linear transformations

1.28 $

Problem: Complement the system of vectors \( e_{1}=\{1,2,3,4\}, \quad e_{2}=\{-1,-3,1,1\} \quad \) to the orthogonal system in \( R^{4} \).

1.7.4 Linear transformations

1.28 $

Problem: Given the matrix of the operator \( A=\left(\begin{array}{lll}2 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1\end{array}\right) \) in basis \( \left(e_{1}, e_{2}, e_{3}\right) \). Find its matrix in basis \( \left(e_{1}^{\prime}, e_{2}^{\prime}, e_{3}{ }^{\prime}\right) \), where \( \left\{\begin{array}{c}e_{1}{ }^{\prime}=e_{1}-e_{2}+e_{3} \\ e_{2}{ }^{\prime}=-e_{1}+e_{2}-2 e_{3} \\ e_{3}{ }^{\prime}=-e_{1}+2 e_{2}+e_{3}\end{array}\right. \).

1.7.5 Linear transformations

1.8 $

Problem: Given the operators \( A x=\left\{x_{2}-x_{3}, x_{1}, x_{1}+x_{3}\right\} \), \( B x=\left\{x_{2}, 2 x_{3}, x_{1}\right\} \). Find the operator \( \left(B^{2}-2 A\right) \).

1.7.6 Linear transformations

1.03 $

Problem: \( \varphi- \) linear operator of rank 1. Prove that at least one of the operators \( \varphi+\varepsilon, \varphi-\varepsilon \) is invertible. ( \( \varepsilon- \) an identity operator).

1.7.7 Linear transformations

2.57 $

Problem: Give an example of a linear operator whose matrix is not diagonal in any basis.

1.7.2 Linear transformations

1.28 $

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