Math Problem Statement

Найти матрицу линейного оператора A_φ в новом базисе {\tilde{e}_i} на основе заданной матрицы A_φ в старом базисе {e_i} и двух заданных базисов.

Solution

To solve this problem and find the matrix of the linear operator A~φ\tilde{A}_\varphi in the new basis {e~i}i=13\{\tilde{e}_i\}_{i=1}^3, we need to follow these steps:


1. Recall the Transformation Formula

The matrix of the operator A~φ\tilde{A}_\varphi in the new basis is related to its matrix AφA_\varphi in the old basis by: A~φ=P1AφP,\tilde{A}_\varphi = P^{-1} A_\varphi P, where:

  • PP is the change of basis matrix from the new basis {e~i}\{\tilde{e}_i\} to the old basis {ei}\{e_i\},
  • P1P^{-1} is the inverse of PP.

2. Find the Change of Basis Matrix PP

To form PP, we express the new basis vectors e~i\tilde{e}_i in terms of the old basis {ei}\{e_i\}. Each column of PP consists of the coordinates of e~i\tilde{e}_i with respect to {e1,e2,e3}\{e_1, e_2, e_3\}. Specifically: [ P = \begin{bmatrix} [\tilde{e}_1]_e & [\tilde{e}_2]_e & [\tilde{e}_3]_e \end{bmatrix}. ]

From the problem: e~1=(125),e~2=(213),e~3=(179).\tilde{e}_1 = \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix}, \quad \tilde{e}_2 = \begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix}, \quad \tilde{e}_3 = \begin{pmatrix} 1 \\ 7 \\ 9 \end{pmatrix}. Thus, PP is: P=(121217539).P = \begin{pmatrix} 1 & -2 & 1 \\ 2 & 1 & 7 \\ 5 & 3 & 9 \end{pmatrix}.


3. Compute P1P^{-1}

We compute the inverse of the 3×33 \times 3 matrix PP: P=(121217539).P = \begin{pmatrix} 1 & -2 & 1 \\ 2 & 1 & 7 \\ 5 & 3 & 9 \end{pmatrix}. The inverse P1P^{-1} can be calculated using standard methods (e.g., row reduction, adjugate formula, or a calculator).


4. Compute A~φ\tilde{A}_\varphi

Once P1P^{-1} is determined, we can compute: A~φ=P1AφP,\tilde{A}_\varphi = P^{-1} A_\varphi P, where AφA_\varphi is the original matrix given as: Aφ=(0012512125).A_\varphi = \begin{pmatrix} 0 & 0 & -1 \\ 2 & -5 & 12 \\ -1 & 2 & -5 \end{pmatrix}.


5. Substitute and Simplify

  • Multiply P1AφPP^{-1} A_\varphi P step by step.
  • Simplify the resulting product to get A~φ\tilde{A}_\varphi.

Final Step

The resulting matrix A~φ\tilde{A}_\varphi is the matrix of the operator in the new basis.

Would you like me to compute the inverse of PP and the final product explicitly? Let me know!


Related Questions

  1. How is the change of basis formula derived?
  2. How do you compute the inverse of a 3×33 \times 3 matrix?
  3. What are the properties of linear operators under a change of basis?
  4. How do eigenvalues of a matrix behave under a change of basis?
  5. Can the matrix PP ever be singular? Why or why not?

Tip:

When changing bases, always verify that the new basis vectors are linearly independent to ensure PP is invertible.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Change of Basis
Matrix Transformation
Linear Operators

Formulas

Change of basis formula: \( \tilde{A}_\varphi = P^{-1} A_\varphi P \)
Matrix multiplication and inversion methods

Theorems

Change of basis theorem for linear transformations
Matrix inverse properties

Suitable Grade Level

Undergraduate