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

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The discussion focuses on proving that the minimal polynomial of a square matrix A is the same as that of its transpose A^T. It begins by defining the minimal polynomial and its properties, specifically that for an eigenvalue λ, (A - λI)^n = 0 while (A - λI)^m ≠ 0 for all m < n. Participants express confusion about how to demonstrate that the same holds for the transpose, suggesting that the relationship (A^n)^T = (A^T)^n may be a key insight. The conversation highlights the connection between the linear independence of vectors generated by A and the implications for the minimal polynomial. Ultimately, the goal is to establish that the minimal polynomials for both A and A^T are identical.
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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 &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?
 
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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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