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lukaszh
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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 :-)
lukaszh said:I'm so sorry, because I've made a mistake. Prove that det(A+I)=1. Thank you again for your reactions.
lukaszh said: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] :-(
A nilpotent matrix is a square matrix where there exists a positive integer k such that A^k = 0, where 0 is the zero matrix. This means that when the matrix is raised to the kth power, it becomes the zero matrix.
The identity matrix (I) is added to the nilpotent matrix (A) to simplify the proof. It allows us to use the identity property of matrix multiplication, which states that multiplying any matrix by the identity matrix will result in the original matrix. This simplifies the proof and allows us to show that det(A+I) = det(A) + det(I) = det(A) + 1 = 0 + 1 = 1.
This is a property of determinants. When a multiple of a row or column is added to another row or column, the determinant remains unchanged. In this case, we are adding the identity matrix (which is a multiple of the original matrix) to the nilpotent matrix, so the determinant remains the same.
Since A is nilpotent, there exists a positive integer k such that A^k = 0. This means that the determinant of A^k is also equal to 0. And since the determinant of a matrix raised to a power is equal to the determinant of the matrix multiplied by itself k times, we can say that det(A^k) = det(A)^k = 0. This means that det(A) = 0.
If det(A+I) = 1, and we have shown that det(A+I) = det(A) + 1 = 0 + 1 = 1, then this means that det(A) = 0. Since we know that det(A) = 0 from the previous steps, this confirms that det(A+I) = 0. Therefore, we have proven that det(A+I) = 0 when A is nilpotent.