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I realised an error in my previous attempt, before writing the question, but think this can help some guys to solve their own multipolynomial problems, if you want to keep it (so they can find it via searchfunction) tell me otherwise I will just delete it all..

Minimal polynomial of $A := \begin{pmatrix} 7 & -2 & 1 \\ -2 & 10 & -2 \\ 1 & -2 & 7 \end{pmatrix}$

I approached it in the following way:

  1. Calculate

$E, A, A^2$

$E := \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $ $A:= \begin{pmatrix} 7 & -2 & 1 \\ -2 & 10 & -2 \\ 1 & -2 & 7 \end{pmatrix}$ $A^2:= \begin{pmatrix} 54 & -36 & 18 \\ -36 & 108 & -36 \\ 18 & 36 & 54 \end{pmatrix}$

  1. Now test, if you can write $A = x\cdot E$ (obviously not possible, because of the all the zeros)

  2. Test, if you can write $A^2$ as $lin(E,A)$ by taking the first column of $E, A, A^2$ and solve the equation: \begin{pmatrix} 1 & 7 & 54 \\ 0 & -2 & -36 \\ 0 & 1 & 18 \end{pmatrix} = \begin{pmatrix} 1 & 0 & -72 \\ 0 & 1 & 18 \\ 0 & 0 & 0 \end{pmatrix} So now we know $A^2 = 18\cdot A-72\cdot E$

  3. So $0 = A^2-(18\cdot A-72\cdot E) \Leftrightarrow 0 = A^2 - 18\cdot A + 72\cdot E$

  4. So the minimalpolynomial should be $x^2-18x +72$

But when I then try to get the zeros, which are $6$ and $12$... Oh wait I made a signerror in my last solution..

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  • $\begingroup$ Don't delete this... fix it and people may learn something. $\endgroup$ – Will Jagy Jan 19 '17 at 2:13
  • $\begingroup$ @Will Jagy It's fixed, I verified with arndt-bruenner.de/mathe/scripts/eigenwert2.htm $\endgroup$ – WhatAMesh Jan 19 '17 at 2:14
  • $\begingroup$ PLEASE do not add "solved" to the title: in this site we mark solved questions by selecting an answer. $\endgroup$ – Mariano Suárez-Álvarez Jan 19 '17 at 2:20
  • $\begingroup$ @Mariano Suárez-Álvarez Should I rather accept an answer that isn't the key to the solution? Is there a tag I should use in this case? $\endgroup$ – WhatAMesh Jan 19 '17 at 2:23
  • $\begingroup$ Well, that would be a solution $\endgroup$ – WhatAMesh Jan 19 '17 at 2:39
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It seems you don't know this. Let $\sigma_1$ be the trace of your matrix. Let $\sigma_3$ be the determinant. Here is the new one, let $\sigma_2$ be the sum of the three determinants of two-by-two submatrices with two elements on the main diagonal. Then the characteristic polynomial is $$ \lambda^3 - \sigma_1 \lambda^2 + \sigma_2 \lambda - \sigma_3. $$ The minimal polynomial must must divide the characteristic polynomial.

I get $$ \lambda^3 - 24 \lambda^2 + 180 \lambda - 432. $$ At the same time, the matrix is symmetric, so there is a basis of (real) eigenvectors. In particular, each row sum is $6,$ so the column vector made up of three $1'$s is an eigenvector with eigenvalue $6.$ I then divided out by $(\lambda - 6) $ and got $\lambda^2 - 18 \lambda + 72,$ which is $(\lambda - 6)(\lambda - 12.)$ The characteristic polynomial is $$ (\lambda - 6)^2 (\lambda - 12). $$ The Jordan normal form is diagonal, as the matrix is symmetric. This means the minimal polynomial is $ (\lambda - 6) (\lambda - 12). $

Let's see, $$A = \begin{pmatrix} 7 & -2 & 1 \\ -2 & 10 & -2 \\ 1 & -2 & 7 \end{pmatrix},$$ $$A - 6 I = \begin{pmatrix} 1 & -2 & 1 \\ -2 & 4 & -2 \\ 1 & -2 & 1 \end{pmatrix},$$ $$A - 12 I = \begin{pmatrix} -5 & -2 & 1 \\ -2 & -2 & -2 \\ 1 & -2 & -5 \end{pmatrix},$$ $$(A-6I)(A - 12 I) = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},$$

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  • $\begingroup$ That's right, im not there yet $\endgroup$ – WhatAMesh Jan 19 '17 at 2:19
  • $\begingroup$ @WhatAMesh, this specific calculation, for 3 by 3, is not always taught. Worthwhile trying it yourself, get confident with it. My students really liked it $\endgroup$ – Will Jagy Jan 19 '17 at 2:24
  • $\begingroup$ I'll look into it! $\endgroup$ – WhatAMesh Jan 19 '17 at 2:24
  • $\begingroup$ @WhatAMesh work out the two by two case first. This should be familiar, just the trace and determinant are required. By the time we get to 4 by 4 it depends too much on special features of the matrix... $\endgroup$ – Will Jagy Jan 19 '17 at 2:28

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