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Hi there I have to calculate the minimal polynomial for the matrix $$ \begin{pmatrix} 2 & 0 & 0 &0 \\ 1 & 3& 2 & 1\\ 0 & -1& 0 &-1 \\ -1& 0 & 0 & 2 \end{pmatrix}$$ In oreder to do that I tried to put its characteristic matrix in the canonical diagonal form, and I got the matrix $$\begin{pmatrix} 1 & 0 & 0 &0 \\ 0 & 1& 0 & 0\\ 0 & 0& -x^2+3x-2 &0\\ 0& 0 & 0 & (x-2)(2-x) \end{pmatrix}$$. From here I see that the minimal polynomial is $ \mu_A(x) = (x-2)(2-x)$, but the problem is that this minimal polynomial dosen't check the Hamilton-Cayley theorem(i.e. $\mu_A(A) = 0$).

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  • $\begingroup$ What you call the canonical diagonal form (I know it as the Smith normal form) requires each diagonal entry to divide then next. In your example this fails for the final two entries, so you did not correctly compute this normal form. $\endgroup$ – Marc van Leeuwen Sep 19 '16 at 19:51
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You seem to have confused some of the concepts. Here, you successfully diagonalized the characteristic matrix. In that form, the characteristic polynomial is the product of the diagonal elements, i.e. $$(-x^2+3x-2)(x-2)(2-x)=(x-2)^3(x-1).$$ Now, the Cayley-Hamilton theorem says that the matrix $A$ should satisfy the characteristic polynomial, i.e. $(x-2)^3(x-1)$. This is true, you can verify it.

Now, the minimal polynomial is by definition the smallest polynomial (i.e. of minimal degree) that $A$ satisfies. It can be shown that the minimal polynomial necessarily divides the characteristic polynomial, and hence it is of the form $(x-2)^m(x-1)^n$ for $m\in\{0,1,2,3\}$ and $n\in\{0,1\}$. You can check that the answer is $\mu_A=(x-2)^2(x-1)$.

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  • $\begingroup$ Also, one can prove that $m,n>0$. Generally, the minimal and characteristic polynomials share the same irreducible factors $\endgroup$ – Guy Sep 16 '16 at 15:45
  • $\begingroup$ I see now... I thought that the minimal polynomial is given by the last invariant factor of the canonical diagonal form (in this case $(x-2)(2-x)$) $\endgroup$ – Raducu Mihai Sep 16 '16 at 20:33
  • $\begingroup$ @RaducuMihai what you say is correct, but requires that you correctly compute the canonical form, which in fact (assuming Spencer computed correctly) has diagonal entries $1$, $1$, $x-2$ and $(x-1)(x-2)^2$. Indeed each factor divides the next, the last one is the minimal polynomial, and the product is the characteristic polynomial. Note also that all these factors are monic polynomials while yours are not. $\endgroup$ – Marc van Leeuwen Sep 19 '16 at 19:57
  • $\begingroup$ Yeah.. got it. Thanks for the help :) $\endgroup$ – Raducu Mihai Sep 20 '16 at 14:05

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