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

In summary, the minimal polynomial for a square matrix and its transpose is the same. This can be shown by using the property (A^n)^T = (A^T)^n and the given conditions for the minimal polynomial.
  • #1
catsarebad
72
0

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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  • #2
Let [itex]\lambda[/itex] be an eigenvalue of [itex]A[/itex] such that
[tex](A - \lambda I)^n = 0[/tex]
but
[tex](A - \lambda I)^{m} \neq 0[/tex]
for every positive integer [itex]m < n[/itex].

Given that, can you show that [itex](A^T - \lambda I)^n = 0[/itex] and that there does not exist a positive integer [itex]m < n[/itex] such that [itex](A^T - \lambda I)^m = 0[/itex]?
 
Last edited:
  • #3
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.
 
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  • #4
pasmith said:
Let [itex]\lambda[/itex] be an eigenvalue of [itex]A[/itex] of geometric multiplicity [itex]n[/itex]. Then
[tex](A - \lambda I)^n = 0[/tex]
but
[tex](A - \lambda I)^{m} \neq 0[/tex]
for every positive integer [itex]m < n[/itex].

Given that, can you show that [itex](A^T - \lambda I)^n = 0[/itex] and that there does not exist a positive integer [itex]m < n[/itex] such that [itex](A^T - \lambda I)^m = 0[/itex]?

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

i assume this is a property
[tex](A - \lambda I)^n = 0[/tex]
but
[tex](A - \lambda I)^{m} \neq 0[/tex]
for every positive integer [itex]m < n[/itex].

but i don't get how showing the next part will help with minimal poly problem.
 
  • #5
catsarebad said:
i'm not sure where we are going with this.

i assume this is a property
[tex](A - \lambda I)^n = 0[/tex]
but
[tex](A - \lambda I)^{m} \neq 0[/tex]
for every positive integer [itex]m < n[/itex].

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

Hint: [itex](A^n)^T = (A^T)^n[/itex]
 

1. What is a minimal polynomial for a square matrix?

A minimal polynomial for a square matrix is the smallest degree monic polynomial that the matrix satisfies. This means that if the matrix A satisfies the polynomial p(x), then there is no other polynomial of smaller degree that A satisfies.

2. How is the minimal polynomial related to a square matrix and its transpose?

The minimal polynomial for a square matrix A and its transpose AT are related in that they have the same degree and the same roots. This means that if A satisfies the minimal polynomial p(x), then AT will also satisfy p(x).

3. Why is it important to show that the minimal polynomial for a square matrix and its transpose are the same?

Showing that the minimal polynomial for a square matrix and its transpose are the same is important because it provides a deeper understanding of the properties of the matrix. It also allows for easier computation and analysis of the matrix, as the minimal polynomial can provide information about the matrix's eigenvalues and other important characteristics.

4. How can one prove that the minimal polynomial for a square matrix and its transpose are the same?

The most common way to prove that the minimal polynomial for a square matrix and its transpose are the same is by using the Cayley-Hamilton theorem. This theorem states that a matrix satisfies its own characteristic polynomial, and since the characteristic polynomial is the same for a matrix and its transpose, this proves that the minimal polynomial is also the same.

5. Are there any exceptions to the rule that the minimal polynomial for a square matrix and its transpose are the same?

Yes, there are some rare cases where the minimal polynomial for a square matrix and its transpose may not be the same. This can occur when the matrix has complex eigenvalues or when the minimal polynomial has repeated roots. However, in most cases, the minimal polynomial will be the same for a matrix and its transpose.

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