Show that minimal poly for a sq matrix and its transpose is the same

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Homework Help Overview

The discussion revolves around proving that the minimal polynomial for a square matrix and its transpose is the same. Participants are exploring the properties of eigenvalues and the implications of the minimal polynomial in relation to matrix transposition.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants attempt to relate the eigenvalues of a matrix to its transpose, questioning how the properties of the minimal polynomial apply in this context. Others express uncertainty about the relevance of certain properties to the minimal polynomial problem.

Discussion Status

The discussion is ongoing, with participants raising questions about the implications of eigenvalue properties and their connection to the minimal polynomial. Hints have been provided, but there is no explicit consensus on the next steps or a clear direction established.

Contextual Notes

Participants note the lack of clarity regarding how certain properties of matrices relate to the minimal polynomial, indicating potential gaps in understanding or assumptions that need to be addressed.

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Homework Statement


show that minimal poly for a sq matrix and its transpose is the same

Homework Equations



The Attempt at a Solution


no clue.
 
Last edited:
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Let \lambda be an eigenvalue of A such that
(A - \lambda I)^n = 0
but
(A - \lambda I)^{m} \neq 0
for every positive integer m < n.

Given that, can you show that (A^T - \lambda I)^n = 0 and that there does not exist a positive integer m < n such that (A^T - \lambda I)^m = 0?
 
Last edited:
catsarebad said:

Homework Statement


show that minimal poly for a sq matrix and its transpose is the same


Homework Equations






The Attempt at a Solution


no clue.

The polynomial ##p(x) = x^m + a_1 x^{m-1} + \cdots + a_{m-1} x + a_m ## is a minimal polynomial for matrix ##A## if and only if ##p(A) x = 0## for all column vectors ##x##, but this is not true for any polynomial of degree < m.

In other words, the vectors ##x, Ax, A^2 x, \ldots, A^{m-1}x## are linearly independent, but ##A^m x## is a linear combination of ##x, Ax, A^2 x, \ldots, A^{m-1}x##; furthermore, this same m and this same linear combination holds for all ##x##.

Basically, this is how some computer algebra packages find minimal polynomials, without finding the eigenvalues first. In fact, if we restrict the field of scalars to the reals a real matrix ##A## may not have (real) eigenvalues at all, but it will always have a real minimal polynomial.
 
Last edited:
pasmith said:
Let \lambda be an eigenvalue of A of geometric multiplicity n. Then
(A - \lambda I)^n = 0
but
(A - \lambda I)^{m} \neq 0
for every positive integer m &lt; n.

Given that, can you show that (A^T - \lambda I)^n = 0 and that there does not exist a positive integer m &lt; n such that (A^T - \lambda I)^m = 0?

i'm not sure where we are going with this.

i assume this is a property
(A - \lambda I)^n = 0
but
(A - \lambda I)^{m} \neq 0
for every positive integer m &lt; n.

but i don't get how showing the next part will help with minimal poly problem.
 
catsarebad said:
i'm not sure where we are going with this.

i assume this is a property
(A - \lambda I)^n = 0
but
(A - \lambda I)^{m} \neq 0
for every positive integer m &lt; n.

but i don't get how showing the next part will help with minimal poly problem.

Hint: (A^n)^T = (A^T)^n
 

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