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

1. Jan 30, 2014

1. The problem statement, all variables and given/known data
show that minimal poly for a sq matrix and its transpose is the same

2. Relevant equations

3. The attempt at a solution
no clue.

Last edited: Jan 30, 2014
2. Jan 30, 2014

### pasmith

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: Jan 30, 2014
3. Jan 30, 2014

### Ray Vickson

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: Jan 30, 2014
4. Jan 30, 2014

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 $m < n$.

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

5. Jan 30, 2014

### pasmith

Hint: $(A^n)^T = (A^T)^n$