Chapter 24: Problem 17
The group of all non-singular \(n \times n\) matrices is known as the general linear group \(G L(n)\) and that with only real elements as \(G L(n, \mathbf{R}) .\) If \(\mathbf{R}^{*}\) denotes the multiplicative group of non-zero real numbers, prove that the mapping \(\Phi\) : \(G L(n, \mathbf{R}) \rightarrow \mathbf{R}^{*}\), defined by \(\Phi(\mathrm{M})=\operatorname{det} \mathrm{M}\), is a homomorphism. Show that the kernel \(\mathcal{K}\) of \(\Phi\) is a subgroup of \(G L(n, \mathbf{R})\). Determine its cosets and show that they themselves form a group.
Short Answer
Step by step solution
Define the Mapping
Show Homomorphism Property
Define the Kernel
Verify Subgroup Property
Determine Cosets
Show Cosets Form a Group
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Nonsingular Matrices
Matrix Determinant
- \( \operatorname{det}(A B) = \operatorname{det}(A) \operatorname{det}(B) \)\
- If \( \operatorname{det}(M) = 0 \), the matrix \( M \) is singular (non-invertible).
- \( \operatorname{det}(M^{-1}) = \operatorname{det}(M)^{-1} \) if \( M \) is nonsingular.
Group Theory
Kernel
- \( \mathcal{K} = \{ M \in GL(n, \mathbf{R}) \mid \operatorname{det}(M) = 1 \} \)
Cosets
- \( A \mathcal{K} \) where \( \operatorname{det}(A) = c \in \mathbf{R}^* \)