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Homework Help: How do you find the minimal polynomial?

  1. Oct 11, 2012 #1
    1. The problem statement, all variables and given/known data

    If we have a transformation matrix [itex]\begin{bmatrix} 1 & 2 & 4 \\0 & 0 & 0 \\0 & 0 & 0 \end{bmatrix}[/itex]

    2. Relevant equations

    3. The attempt at a solution

    I found the characteristic polynomial of this matrix: [itex] x^3 - x^2 [/itex] = [itex] x^2(x-1) [/itex]...can anybody please help me?

    Thanks in advance
  2. jcsd
  3. Oct 11, 2012 #2

    Simon Bridge

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  4. Oct 11, 2012 #3
    I do know that the minimal polynomial divides the characteristic polynomial. But there is nothing else about how to find it in my notes. But thanks, I will look at the link that you gave me...
  5. Oct 11, 2012 #4


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    Surely there is a definition? The "minimal polynomial" of a linear transformation is the polynomial of lowest degree that is made equal to 0 by the linear transformation. Yes, it must divide the "characteristic polynomial". And the only divisors of [itex]x^2(x- 1)[/itex] are [itex]x[/itex], [itex]x^2[/itex], [itex]x- 1[/itex], [itex]x(x- 1)[/itex], and [itex]x^2(x- 1)[/itex]. Which of does A make 0?
  6. Oct 11, 2012 #5

    Ray Vickson

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    You could do what Maple does, which is to regard the identity matrix I and the matrices A, A^2, A^3,... as being spread out into n^2-dimensional vectors, then use row operations to find the least m such that I,A,A^2, ... , A^m are linearly dependent, and to find the coefficients involved.

    However, in your case that is overkill. Just follow the previous suggestions, and test the polynomials x^2 and x(1-x) to see if either of them annihilate A. If not, minimal = characteristic.

  7. Oct 11, 2012 #6
    By, "A", you mean the transformation matrix, right?
  8. Oct 11, 2012 #7
    Thanks a lot. I think I was able to do it for this matrix...but I have another matrix that I'm a bit confused about...

    [itex]\begin{bmatrix} 1 & cos( \theta) \\0 & sin( \theta) \end{bmatrix}[/itex] where theta is between zero and 2pi.

    This is what the professor told us:

    If theta is not equal to pi, then the minimal polynomial is equal to the characteristic polynomial. If theta equals pi, then the minimal polynomial is the negative of the identity. I found the characteristic polynomial to be...[itex]f(x)= x^2 - x - x sin( \theta) + sin( \theta)[/itex]...but now, I have theta in the equation. So how will that work? I won't be able to substitute my matrix for x and get zero when theta is in there...
  9. Oct 11, 2012 #8

    Ray Vickson

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    Yes, of course. You have a matrix, and you need sometimes to give it a name.

  10. Oct 11, 2012 #9
    Can you expand on what you have said about that maple thing? Because our professor said something similar...but I'm not sure I understood him correctly...

    He told us to use this matrix:

    [itex]\begin{bmatrix} 1 & 2 & 4 & 8\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{bmatrix}[/itex]

    to make it linearly dependent (this question actually has many parts, so the matrix I gave is related to this one...except that this one is linearly dependent of course...this is the least one that is linearly dependent, by the way). So can you explain the maple method too? If you don't mind...
  11. Oct 11, 2012 #10

    Ray Vickson

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    This won't work: is is a 3×4 matrix, so you cannot even square it. If we drop column 4 we obtain a 3×3 matrix A, but it is too easy to show what is happening, since it satisfies A2 = A, so has minimum polynomial x2- x. Let's try instead the matrix
    [tex]B = \begin{bmatrix}1&2&3\\1&0&0\\0&1&1 \end{bmatrix},\\
    \text{ giving }\\
    B^2 = \begin{bmatrix}3&5&6\\1&2&3\\1&1&1\end{bmatrix} \\
    B^3 = \begin{bmatrix} 8&12&15\\3&5&6\\2&3&4\end{bmatrix}
    Form the vectors
    [tex] I \leftrightarrow v_0 = \begin{bmatrix}1&0&0&0&1&0&0&0&1\end{bmatrix}\\
    A \leftrightarrow v_1 = \begin{bmatrix}1&2&3&1&0&0&0&1&1\end{bmatrix}\\
    A^2 \leftrightarrow v_2 = \begin{bmatrix}3&5&6&1&2&3&1&1&1\end{bmatrix}\\
    A^3 \leftrightarrow v_3 = \begin{bmatrix}8&12&15&3&5&6&2&3&4\end{bmatrix}
    Now start doing row-operations on v0,v1,v2,v3:
    [tex] v_1' = v_1-v_0 = \begin{bmatrix}0&2&3&1&-1&0&9&1&0\end{bmatrix}\\
    v_2' = v_2 -3v_0 = \begin{bmatrix}0&5&6&1&-1&3&1&1&-2\end{bmatrix}\\
    v_3' = v_3 - 8v_0 = \begin{bmatrix}0&12&15&3&-3&6&2&3&-4\end{bmatrix}\\
    Since v_1' ≠ 0 and v_2' ≠ 0 we are not yet done with row-reduction. Now "zero out" the second components of v_2' and v_3', by subtracting multiples of v_1':
    [tex] v_2'' = v_2' - (5/2)v_1' = \begin{bmatrix}0&0&-3/2&-3/2&3/2&3&1&-3/2&2\end{bmatrix}\\
    v_3'' = v_3' - 6v_1' = \begin{bmatrix}0&0&-3&-3&3&6&2&-3&-4\end{bmatrix}
    Since v_2'' ≠ 0 the minimal polynomial is of degree 3 (so is the characteristic polynomial). If we want to, we can even determine characteristic polynomial by continuing the process:
    v_3''' = v_3'' - [(-3)/(-3/2)]v_2'' = v_3'' -2 v_2'' =
    Thus, we have
    [tex] 0 = v_3''' = v_3 - 2v_2 - v_1 - v_0 \leftrightarrow A^3 - 2A^2 - A - I, [/tex]
    so the characteristic polynomial is
    [tex] p(x) = x^3 - 2x^2 - x - 1.[/tex]

  12. Oct 14, 2012 #11
    Thanks a lot :)
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