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(True or false):
Suppose $A$ is an $n \times n$ matrix and $A^{k} = 0$ for some k. Then the characteristic polynomial is $x^n$

I am inclined to believe this is true since if $x^{k} = 0$ then the minimal polynomial divides $x^k$ thus implying that the minimal polynomial has the form of $x^a$ for some $a$. Which implies that the characteristic polynomial has the form of $x^b$ for some $b$ since the minimal polynomial divides the characteristic polynomial. I know that the characteristic polynomial for an $n \times n$ matrix will have the form of $x^n -trace(A)x^{n-1} + ....+(-1^n)det(A)$
So if the characteristic polynomial has the form $x^b$ then for an $n \times n$ matrix it will have the form $x^n$

Another question is (True or false)
Suppose $A^k = 0$ for some $k$. Then the minimal polynomial of A is $x^k$

I am inclined to believe this is false since the question doesnt really state weather $k < n$ or $k > n$ or if $k$ is the first number which sets $A$ to zero. Thus there can be an $a < k$ where $A^a = 0$

Is my reasoning sound for both examples?

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    $\begingroup$ The characteristic polynomial of any square matrix $\;n\times n\;$ over any field is always of degree $\;n\;$ . $\endgroup$ – DonAntonio May 12 '16 at 22:28
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Regarding your first question, the answer is yes. Since $A^k = 0$, the minimal polynomial must divide $x^k$ so it is of the form $x^i$ for $1 \leq i \leq k$. While the minimal polynomial divides the characteristic polynomial by the Cayley-Hamilton theorem, it is not enough to deduce that the characteristic polynomial must be $x^n$ (it can be a priori, for example, $x^k(x-1)^{n-k}$). However, the minimal polynomial and the characteristic polynomial have the same irreducible factors and this implies that the characteristic polynomial must be $x^n$.

Regarding your second question, you are correct. If $A = 0_{n \times n}$ then $A^k = 0$ for any $k \in \mathbb{N}$ but the minimal polynomial of $A$ is $x$.

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  • $\begingroup$ Thanks for the correction! $\endgroup$ – levap May 12 '16 at 23:06

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