Math Problem Statement
Найти в стандартном базисе матрицу оператора проектирования пространства R⁴ на собственное пространство оператора φ ∈ Hom(R⁴, R⁴), соответствующее собственному числу λ = 6. Оператор φ задан своей матрицей в стандартном базисе:
A_φ = ( 90 -42 42 42 ; 56 -22 28 28 ; 0 0 6 0 ; -140 70 -70 -64 )
Solution
To solve the problem of finding the matrix of the projection operator of the given space onto the eigenspace corresponding to eigenvalue , we proceed as follows:
Step 1: Matrix Analysis
The matrix of the operator is:
\begin{pmatrix} 90 & -42 & 42 & 42 \\ 56 & -22 & 28 & 28 \\ 0 & 0 & 6 & 0 \\ -140 & 70 & -70 & -64 \end{pmatrix}.$$ The eigenvalue $$\lambda = 6$$ means we need to **extract the eigenspace corresponding to $$\lambda = 6$$**. --- ### Step 2: Projection Operator Formula The projection matrix $$P$$ onto the eigenspace $$V_{\lambda}$$ of $$\lambda$$ can be expressed as: $$P = v_i v_i^T,$$ where $$v_i$$ are eigenvectors corresponding to $$\lambda = 6$$. However, in this case, for a matrix projection operator to a known eigenspace, we can compute it directly using: $$P = \frac{1}{\lambda} A_{\varphi}^2$$ reAsk a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Eigenvalues and Eigenvectors
Projection Operators
Formulas
Projection operator: P = v_i v_i^T for an eigenspace
Eigenvalue equation: A v = λ v
Theorems
Spectral Decomposition Theorem
Eigenvector Properties
Suitable Grade Level
Undergraduate Level (Linear Algebra Course)