Linear Mathematics | Part IB, 2003

Define the determinant det(A)\operatorname{det}(A) of an n×nn \times n complex matrix A. Let A1,,AnA_{1}, \ldots, A_{n} be the columns of AA, let σ\sigma be a permutation of {1,,n}\{1, \ldots, n\} and let AσA^{\sigma} be the matrix whose columns are Aσ(1),,Aσ(n)A_{\sigma(1)}, \ldots, A_{\sigma(n)}. Prove from your definition of determinant that det(Aσ)=ϵ(σ)det(A)\operatorname{det}\left(A^{\sigma}\right)=\epsilon(\sigma) \operatorname{det}(A), where ϵ(σ)\epsilon(\sigma) is the sign of the permutation σ\sigma. Prove also that det(A)=det(At).\operatorname{det}(A)=\operatorname{det}\left(A^{t}\right) .

Define the adjugate matrix adj(A)\operatorname{adj}(A) and prove from your definitions\operatorname{definitions} that Aadj(A)=A \operatorname{adj}(A)= adj(A)A=det(A)I\operatorname{adj}(A) A=\operatorname{det}(A) I, where II is the identity matrix. Hence or otherwise, prove that if det(A)0\operatorname{det}(A) \neq 0, then AA is invertible.

Let CC and DD be real n×nn \times n matrices such that the complex matrix C+iDC+i D is invertible. By considering det(C+λD)\operatorname{det}(C+\lambda D) as a function of λ\lambda or otherwise, prove that there exists a real number λ\lambda such that C+λDC+\lambda D is invertible. [You may assume that if a matrix AA is invertible, then det(A)0\operatorname{det}(A) \neq 0.]

Deduce that if two real matrices AA and BB are such that there exists an invertible complex matrix PP with P1AP=BP^{-1} A P=B, then there exists an invertible real matrix QQ such that Q1AQ=BQ^{-1} A Q=B.

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