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Math Help - Matrix Polynomial

  1. #1
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    Matrix Polynomial

    So here is a question I have been working on, any suggestions would be great.

    If B is an n x n matrix with enteries in F, then prove that there is a nonzero polynomial p \in F[t] which has B as a root.

    So here is what I have so far. There must exist a_0,..., a_r not all 0 such that


    p(t) = a_0 + a_1t + ... + a_rt^r

    p(B) = a_0I + a_1B + ... + a_rB^r = 0


    So pretty much from this point we just need to show that the set I,B,B^2,...,B^r is linearly dependent. Any advice on how to go about doing this?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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  3. #3
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    <br /> <br />
B=SDS^{-1}=S\begin{bmatrix} \ddots & & \\<br />
 & \lambda_{i} & \\<br />
 & & \ddots \\<br />
 \end{bmatrix}S^{-1}~,~<br />
 p(B)=a_0I + a_1B + \cdots + a_rB^r <br />

    <br />
 =S\begin{bmatrix}<br />
 \ddots & & \\<br />
 & a_0 + a_1\lambda_{i} + \cdots + a_r\lambda_{i}^{r} & \\<br />
 & & \ddots \\<br />
 \end{bmatrix}S^{-1}=0<br />

    <br />
 \rightarrow a_0 + a_1\lambda_{i} + \cdots + a_r\lambda_{i}^{r} = 0 \\<br />
 \rightarrow p(t)=k\cdot \det(tI_n-B)~,~k\in F<br /> <br />
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by math2009 View Post
    <br /> <br />
B=SDS^{-1}=S\begin{bmatrix} \ddots & & \\<br />
 & \lambda_{i} & \\<br />
 & & \ddots \\<br />
 \end{bmatrix}S^{-1}~,~<br />
 p(B)=a_0I + a_1B + \cdots + a_rB^r <br />

    <br />
 =S\begin{bmatrix}<br />
 \ddots & & \\<br />
 & a_0 + a_1\lambda_{i} + \cdots + a_r\lambda_{i}^{r} & \\<br />
 & & \ddots \\<br />
 \end{bmatrix}S^{-1}=0<br />

    <br />
 \rightarrow a_0 + a_1\lambda_{i} + \cdots + a_r\lambda_{i}^{r} = 0 \\<br />
 \rightarrow p(t)=k\cdot \det(tI_n-B)~,~k\in F<br /> <br />
    You seem to be assuming that F is algebraically closed.
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  5. #5
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    Thank you both very much for your help, however I am still a bit confused. We have not studied determinants (except briefly for 2x2 matrices). Would it be possible for you to give a short explanation of your proof?

    The wikipedia page dealt a lot with determinants, eigenvectors and eigenvalues, all of which we have not yet been taught, so I found the proofs hard to follow.

    Thanks!
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  6. #6
    MHF Contributor Bruno J.'s Avatar
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    Okay; let's do it without the machinery.

    Recall that F_{n\times n} is a vector space of degree n^2 over F. Thus any n^2+1 elements of F_{n\times n} are linearly dependent. So consider the set \{B, B^2, \dots, B^{n^2+1}\}... I think I've said enough!

    By the way, the Latex symbol for \times is \times.
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  7. #7
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    Thank you very much, I got it!

    And thanks for the tip, im pretty new to latex.
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