Nilpotent Matrices: Invertibility and Transpose Proof

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SUMMARY

Nilpotent matrices are not invertible, as demonstrated by the fact that their determinant is always zero, which confirms that they cannot have an inverse. The transpose of a nilpotent matrix remains nilpotent, as shown with the example matrix A = [0 1; 0 0], where (A)^τ = [0 0; 1 0] also satisfies the nilpotent condition. Furthermore, the matrix I - A is invertible for nilpotent matrices, which can be proven by assuming the contrary and showing that it leads to a contradiction regarding the determinant.

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Homework Statement


A. Are nilpotent matrices invertible ? Prove your answer.

B. If A is nilpotent, what can you say about (A)^τ ? Prove your answer.

C. If A is nilpotent, show I-A is invertible.



Homework Equations



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The Attempt at a Solution



A. I know invertible matrix are - AB = BA = I

B. I took a nilpotent matrix
A = [ 0 1
0 0 ]
Its transpose is -
(A)^τ = [ 0 0
1 0 ]
And the transpose is still a nilpotent matrix because
(A^τ)^2 = [ 0 0
0 0 ]
But I don't know if its true for all and it says prove your answer.

C. No idea
 
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A matrix is invertible if and only if it has non-zero determinant. What can you say about the determinant of a nilpotent matrix?
C. You could prove this by assuming it is false i.e. I-A is not invertible and then proceeding.
 
Oster said:
A matrix is invertible if and only if it has non-zero determinant. What can you say about the determinant of a nilpotent matrix?
C. You could prove this by assuming it is false i.e. I-A is not invertible and then proceeding.

A nilpotent matrix has determinant 0 since its diagonals are all 0 (Eigen values are 0). So the inverse would be 0 too .
 

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