0
$\begingroup$

So I have a $3\times 3$ matrix, lets say...

$$ A= \begin{pmatrix} 1 &1 &0 \\ 0 &1 &1 \\ 0 &0 &1 \end{pmatrix}.$$

If I calculate the characteristic polynomial, I get $(1-X)^3=0$. Now, the possible minimal polynomials are $(1-x)$ and $(1-x)^3$ (if I'm not wrong). Both can divide $(1-x)^3$ without any rest. Well, now the question: how can I choose between the two? If I'm not wrong, in this case, I can just take $(1-x)$ instead of $(1-x)^3$ because what I want is the polynomial with smaller degree. Is this correct?

Thanks in advance.

$\endgroup$
2
$\begingroup$

Well, the minimal polynomial could be $X - 1$ or $(X -1 )^{2}$ or $(X -1)^{3}$.

Just compute $(X -1 )^{2}$ on the given matrix $A$. To do that, note that $A - 1$ has a very simple form.

$\endgroup$
  • $\begingroup$ No sorry, what you mean for compute $(x-1)^2$ on A? $\endgroup$ – Osvaldo Paniccia Jul 7 '16 at 13:35
  • $\begingroup$ Substitute $A$ in $(X - 1)^{2}$. To check whether $(X - 1)^{2}$ or $X - 1$ is the minimal polynomial. $\endgroup$ – Andreas Caranti Jul 7 '16 at 13:35
  • 1
    $\begingroup$ @OsvaldoPaniccia What's the minimal polynomial, then, if you do not know the meaning of "evaluating a polynomial in a matrix"? $\endgroup$ – user228113 Jul 7 '16 at 13:36
  • $\begingroup$ To help with the calculation, if $e_{i}$ is the standard basis (column) vector, note that $A-1 : e_{3} \mapsto e_{2} \mapsto e_{1} \mapsto 0$. So $(A-1)^{2} : e_{3} \mapsto \dots$. $\endgroup$ – Andreas Caranti Jul 7 '16 at 13:38
  • $\begingroup$ So if im not wrong i need to calculate: $\begin{pmatrix} 0 &0 &-1 \\ -1 &0 &0 \\ -1 &-1 &0 \end{pmatrix}$ $\cdot$ $\begin{pmatrix} 0 &0 &-1 \\ -1 &0 &0 \\ -1 &-1 &0 \end{pmatrix}$ $\endgroup$ – Osvaldo Paniccia Jul 7 '16 at 13:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.