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We have the matrix $A= \begin{pmatrix} 0 &2 &2 \\ 2& 0 &2 \\ 2& 2 & 0 \end{pmatrix}$, then one of the following is true:

  1. $f_A(x)=m_A(x) $

  2. The matrix $C=A^4-4A^2+5I$ isn't diagnolizable over $\mathbb R$

  3. For the matrix $B=A^5+6A^3+A^2-I$ we have $f_B(x)\neq m_B(x)$

  4. For the matrix $B=A^5+6A^3+A^2-I$ we have $f_B(x)$ is a simple polynomial over $\mathbb C$

EDIT: I get that $f_A(x) = x(2+x)(2-x)$ and it looks like I can't zero the matrix with any smaller combination than $f_A$, so it's maybe true.

For 2 I get that eigenvalues of $A$ aren't zeroing $x^4-4x^2=x^2(x^2-4)=-5$, I'm not sure what to make of it though.

For 3, we have $f_B(x) = x^2(x^3+6x+1)-1$ (?) that can't be a minimal polynomial so that's maybe true.

For 4, I don't think it's true for the same reason as 3.

Notes: $f_A(x)$ is the characteristic polynomial of $A$, $m_A(x)$ is the minimal polynomial of A, a "simple" polynomial: $x(x+1)$, not "simple" poly': $x^2(x+3)$

shinzou
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1 Answers1

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The matrix $A$ is real and symmetric, so it must be diagonalisable. Therefore the minimal polynomial (supposedly called $m_A$) has simple roots.

The matrix $A+2I$ has rank $1$ and trace $6$, so its characteristic polynomial is $X^2(X-6)$, and the characteristic polynomial of $A$ (supposedly called $f_A$) is obtained from it by substituting $X+2$ for $X$: $f_A=(X+2)^2(X-4)$ (equivalently, you can see directly that $\lambda=-2$ is an eigenvalue with geometric multiplicity$~2$, and the remaining eigenvalue must be $4$ to make their sum equal to $0$, the trace of$~A$). Since this has a double root, point 1 must fail. In fact $m_A=(X+2)(X-4)$ is now forced.

Any polynomial of a diagonalisable matrix is diagonalisable (on the same basis of eigenvectors), so point 2 must be false as well.

In fact, there is a eigenspace for $A$ of dimension$~2$ (for eigenvalue $\lambda=-2$ of $A$), and this is contained in the eigenspace of any polynomial $P[A]$ of$~A$ (for the eigenvalue $\lambda=P[-2]$ of $P[A]$), so $f_{P[A]}$ always has a multiple root, and can never be equal to $m_{P[A]}$ (for the same reason as for $f_A$ and $m_A$ above). So point 3 is true and point 4 is false, independently of the precise polynomial $P=X^5+6X^3+X^2-1$ used there.

Community
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Marc van Leeuwen
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  • Can we actually find $f_B$ from the given? And if so, how? – shinzou Feb 05 '16 at 10:35
  • Yes, $P[A]$ acts on the $2$-dimensional eigenspace of$~A$ for $\lambda=-2$ by the scalar $P[-2]=-77$ and on the $1$-dimensional eigenspace of$~A$ for $\lambda=4$ by the scalar $P[4]=1423$, so $f_B=(X+77)^2(X-1423)$. – Marc van Leeuwen Feb 05 '16 at 12:42
  • What is $P[x]$? Also, by multiple root you mean the $2$ from $f_B=(X+77)^2(X-1423)$? How can you know for sure that $(X+77)(X-1423)$ zero $B$? – shinzou Feb 05 '16 at 20:37
  • Also, is there another way to find $f_B$ and $m_B$? – shinzou Feb 05 '16 at 21:09
  • $P[x]$ means the result of substituting $x$ for $X$ into the polynomial $P$. For instance, with $P=X^5+6X^3+X^2-1$ and $x=-2$ one has $(X^5+6X^3+X^2-1)[-2]=(-2)^5+6(-2)^3+(-2)^2-1=-77$, which is the scalar by which $B=P[A]$ acts on eigenvectors for $\lambda=-2$ of $A$. Also "multiple" root is the opposite of "simple" root. I used it at some point instead of "double" because I did not know yet (and did not care) whether it might actually be a triple root (it might happen that $P[-2]=P[4]$ for a certain $P$). – Marc van Leeuwen Feb 06 '16 at 10:51
  • That $(X+77)(X-1432)$ annihilates $B$ follows from the fact that $(B+77I)(B-1432I)$ acts as $0$ on each of a basis of eigenvectors for$~A$. – Marc van Leeuwen Feb 06 '16 at 11:12
  • I understand now, thank you. – shinzou Feb 06 '16 at 11:14