Proving the Nilpotency of Square Triangular Matrices with Zero Diagonal Entries

In summary, to prove that any square triangular matrix with each diagonal entry equal to zero is nilpotent, one can demonstrate that as the matrix is raised to higher powers, the zeros in the matrix move closer to the top-right corner. This suggests proving a stronger result that shows the product of two matrices with specific zero entries will also result in a matrix with zero entries. Additionally, the eigenvalues of a nilpotent operator are all 0, and this is not an assumption but rather a fact that can be used to prove this statement.
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Homework Statement


Prove that any square triangular matrix with each diagonal entry equal to zero is nilpotent

The Attempt at a Solution


Drawing out the matrix and multiplying seems a little tedious. Perhaps there is a better way?
Is there another way to do this without assuming that the eigenvalues of a nilpotent operator are all 0?

Thanks for your help!
 
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This might not be the easiest way, but trying a few examples shows that as you raise the matrix to higher powers, the zeros creep up towards the top-right corner one space at a time. This suggests trying to prove a stronger result, that if Aij=0 for i>j-k and Bij=0 for i>j-l, then (AB)ij=0 for i>j-k-l (or something like that).
 
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Since it is triangular with zero diagonal you know the eigenvalues are all 0, thus you know the min poly is x^n for some n. Why are you 'assuming' that the eigenvalues of a nilpotent operator are all 0? It is clearly true (over a field), and isn't important for this question, really (you state you don't want to assume nilpotent implies all e-values 0, but we actually need all e-values 0 implies nilpotent).
 
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FAQ: Proving the Nilpotency of Square Triangular Matrices with Zero Diagonal Entries

What is a nilpotent operator?

A nilpotent operator is a linear transformation on a vector space that, when applied repeatedly, eventually results in the zero vector. In other words, there exists some positive integer n such that the nth power of the operator is the zero operator.

How can I identify a nilpotent operator?

A linear operator is nilpotent if and only if its characteristic polynomial is x^n, where n is the dimension of the vector space on which it acts. In matrix form, a nilpotent operator will have all zeros on and above its main diagonal.

What is the significance of nilpotent operators?

Nilpotent operators are important in the study of linear algebra and functional analysis, as they provide a way to decompose more complex operators into simpler, nilpotent ones. They also have applications in areas such as differential equations and quantum mechanics.

Can nilpotent operators have non-zero eigenvalues?

No, by definition, a nilpotent operator has a characteristic polynomial of x^n, which means that all of its eigenvalues are equal to 0. This also means that the only eigenvector of a nilpotent operator is the zero vector.

How are nilpotent operators related to nilpotent matrices?

A nilpotent matrix is a matrix representation of a nilpotent operator. In other words, a nilpotent operator can be expressed as a nilpotent matrix when a basis for the vector space is chosen. However, not all nilpotent matrices represent nilpotent operators, as the choice of basis may affect the nilpotency of the matrix.

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