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Hello,
I don't want how to prove: Matrix A is nilpotent, so A^k=0. Prove that det(A+I)=0.
Thank you so much :-)
I don't want how to prove: Matrix A is nilpotent, so A^k=0. Prove that det(A+I)=0.
Thank you so much :-)
I'm so sorry, because I've made a mistake. Prove that det(A+I)=1. Thank you again for your reactions.
Hello,
so, I know that
[tex]A=S\Lambda S^{-1}[/tex]
Eigenvalues of A are [tex]\{\lambda_1,\lambda_2,\cdots,\lambda_n\}[/tex] for [tex]n\times n[/tex] matrix. If I add to both sides identity, then
[tex]A+I=S\Lambda S^{-1}+I=S\Lambda S^{-1}+SS^{-1}=S(\Lambda+I)S^{-1}[/tex]
Its determinant is
[tex]\mathrm{det}(A+I)=\mathrm{det}S(\Lambda+I)S^{-1}=\mathrm{det}(\Lambda+I)=\prod_{j=1}^{n}(\lambda_j+1)[/tex]
I don't know, how to utilize fact [tex]A^k=0[/tex] :-(