Consider $$A = \begin{bmatrix} 0 &4&1&-2\\-1&4&0&-1\\0&0&1&0 \\-1&3&0&0 \end{bmatrix}$$

Find the minimal polynomial of $A$ .

Is there any easy/tricky method to find minimal polynomial of this matrix so that I can save my time in examination hall?

Any Hints/solution

Thank you!

  • $\begingroup$ Nop...I don't think so. Just to evaluate the corresponding determinant for the characteristic polynomial and etc. by the usual methods. In this case, pivoting at the entry $3-3$ simplifies things a little...but not that much. $\endgroup$ – DonAntonio Nov 23 '18 at 0:03
  • $\begingroup$ If that's what you're asking, there is no general method/trick you can use to always find the polynomial without resorting to the usual calculations. Yes there are very special matrices where some tricks may be applied, but this is the rare and improbable case. $\endgroup$ – YiFan Nov 23 '18 at 0:14

I think there is no easy method to find the minimal polynomial of this matrix, but rather the following idea might help:

To find the characteristic polynomial for this matrix is easy, since the third row has three zeros. so expanding the determinant of $A-xI$ in the third row to see, $$\rho_A(x)=(1-x).\begin{vmatrix} -x &4&-2\\-1&4-x&-1\\-1&3&-x\end{vmatrix}=-(1-x)(x-1)^2(x-2)=(x-1)^3(x-2)$$ so $\rho_A(x)=0$ if $(x-1)^3=0$ or $(x-2)=0$ and hence $$\sigma(A)=\{1,2\}$$

But minimal polynomial $m_A(x)$ and characteristic polynomial $\rho_A(x)$ have same irreducible factors, so $$m_A(x)\in \Big\{(x-1)(x-2),(x-1)^2(x-2),(x-1)^3(x-2)\Big\}$$

Note that $m_A(x)=\rho_A(x) \iff \dim(\text{each eigenspace})=1$. This is not the case in this matrix , so $m_A(x) \neq (x-1)^3(x-2)=\rho_A(x)$

Now if $m_A(x)=(x-1)(x-2)$, then $(A-I)(A-2I)=0$ which means $$\text{Im}(A-2I) \subset \text{ker}(A-I)$$ so $\text{rank}(A-2I)=3 \leq \text{null}(A-I)=2$, which is false

Hence $$m_A(x)=(x-1)^2(x-2)\;[\text{check!}]$$

  • $\begingroup$ How did you conclude that $\text{dim}(\text{each eigenspace})\ne1$? $\endgroup$ – Klaas van Aarsen Nov 24 '18 at 20:22
  • $\begingroup$ Here 1 is an eigen value with multiplicity 3 and it's corresponding eigen space has dimension not equal to 1.thats why I conclude this. $\endgroup$ – Chinnapparaj R Nov 24 '18 at 20:31
  • $\begingroup$ I just take the negation of the mentioned if and only if result $\endgroup$ – Chinnapparaj R Nov 24 '18 at 20:36
  • $\begingroup$ Let me rephrase that, if eigenvalue 1 has an eigenspace with dimension 3, then the minimal polynomial is equal to the characteristic polynomial, isn't it? That's not the case here, but I don't understand how it might be ruled out. $\endgroup$ – Klaas van Aarsen Nov 24 '18 at 21:12

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