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Question Let X be a square matrix $d(X)=dim (span${$X^i|i\ge 0$})

a. Prove that $d(X) = deg (m_X)-1$

b. prove that $d\begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix} \le d(A)+d(B) \iff m_a , m_b $ are coprime.

Thoughts: a. I think the way is induction. if $deg(m_X)=2$ then $m_X=aX^2+bX+cI$ and so $X^2$ can be expressed as a linear combination of X and I and this implies that d(X)=2. This is contrary to what we need to prove... I found a similar question here Dimension of a span of matrix powers that supports my thought. I need to understand what is the difference between the questions. Is my question problematic?

b. I don't even know where to start this one...

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jreing
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1 Answers1

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$\newcommand{\span}{\mathrm{span}}\newcommand{\lcm}{\mathrm{lcm}}$To expand on my comments, I believe the correct form of part (a) is

$d(X) = \deg (m_X)$, where $d(X)=\dim (\span \{I, X^i \mid i> 0 \})$.

Also, the correct form of part (b) should be

One has $$\tag{ineq} d\left(\begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}\right) \le d(A)+d(B), $$ and equality holds iff $\gcd(m_{A}, m_{B}) = 1$.

To prove this, note that if $$ C = \begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}, $$ then $m_{C} = \lcm(m_{A}, m_{B})$. This is because $f(C) = 0$ iff and only if $f(A) = 0 = f(B)$ if and only if $m_{A} \mid f$ and $m_{B} \mid f$. This proves (ineq), as $\lcm(m_{A}, m_{B})$ divides $m_{A} m_{B}$.

Because of the formula $$\gcd(m_{A}, m_{B}) \cdot \lcm(m_{A}, m_{B}) = m_{A} m_{B},$$ equality holds in (ineq) if and only if $\gcd(m_{A}, m_{B}) = 1$.

Andreas Caranti
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