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Thread: Dimension of matrix spanning set

  1. #1
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    Dimension of matrix spanning set

    Studying for my final in linear algebra and came across this:

    Problem: Let V be the vector space of all real n x n matrices over R. Let A be an element of V, and let W be the subspace of V spanned by $\displaystyle I, A, A^2, A^3 ...$. Prove that $\displaystyle dim(W) \leq n$.

    Thoughts: The statement seems intuitive, but I'm not sure exactly what theorem/properties it results from. I know that a subspace cannot have dimension larger than its finite parent space, so $\displaystyle dim(W) \leq dim(V) = n$. However, my professor said that I should try to prove it by having it follow from the dimension of a matrix after raising it to higher powers; i.e. not from the obvious theorem.
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  2. #2
    Super Member girdav's Avatar
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    Re: Dimension of matrix spanning set

    The dimension of $\displaystyle V$ is $\displaystyle n^2$, not $\displaystyle n$.
    Do you know Cayley-Hamilton theorem?
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  3. #3
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    Re: Dimension of matrix spanning set

    We touched on it in class, but have not gone through it rigorously. I'll read up on it now. Does this mean that there is a typo in the question, i.e. $\displaystyle dim(W) \leq n^2 $ then?
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    Super Member girdav's Avatar
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    Re: Dimension of matrix spanning set

    No, the dimension of $\displaystyle W$ is at most $\displaystyle n$, since we can write $\displaystyle A^n$ as $\displaystyle a_0\cdot I+a_1\cdot A+\ldots+a_{n-1}A^{n-1}$
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  5. #5
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    Re: Dimension of matrix spanning set

    I read up on Cayley Hamilton, but am a little shaky as far as its implications. According to the theorem, the characteristic equation of any n x n matrix, A, satisfies $\displaystyle A^n + c_1 A^{n-1} c_2 +...+ c_{n-2} A + c_{n-1} I_n = 0 $, where $\displaystyle c_i$ are scalars, and $\displaystyle I_n$ is the identity matrix. But that creates, at most, $\displaystyle n^2 $ equations.

    Is the dimension related to the maximum number of solutions to this system? I don't see how to show that it would be at most n.
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  6. #6
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    Re: Dimension of matrix spanning set

    Does anyone have further insight to this question? I still haven't been able to solve it or relate it back to Cayley Hamilton.
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  7. #7
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    Re: Dimension of matrix spanning set

    if A satisfies $\displaystyle A^n + c_1 A^{n-1} + c_2 A^{n-2}+...+ c_{n-2} A + c_{n-1} I_n = 0 $,

    that is, if $\displaystyle p(x) = x^n + c_1x^{n-1} + c_2x^{n-2} + \dots + c_{n-1}$, so that p(A) = 0, then

    $\displaystyle A^n = -(c_1 A^{n-1} c_2 +...+ c_{n-2} A + c_{n-1} I_n)$ which means the n-th power (and thus any higher power) of A

    is a linear combination of $\displaystyle \{I_n,A,A^2,\dots,A^{n-1}\}$.

    but this means that this set spans W, so dim(W) ≤ n.

    and the characteristic polynomial of A, det(xI - A), is just such a polynomial (of degree n).

    the cayley-hamilton theorem says: if p(x) = det(xI - A), then p(A) = 0.
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Dimension of matrix spanning set

    We can also assure that if $\displaystyle \mu$ is the minimal polynomial of $\displaystyle A$ and $\displaystyle m=\textrm{deg}\;(\mu)$ then, $\displaystyle \dim V=m$ and $\displaystyle B=\{I,A,A^2,\ldots, A^{m-1}\}$ is a basis of $\displaystyle V$ .
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  9. #9
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    Re: Dimension of matrix spanning set

    furthermore, μ(x) divides det(xI - A).
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Dimension of matrix spanning set

    Furthermore, $\displaystyle \mu (x)$ divides any annihilator polynomial of $\displaystyle A$ .
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  11. #11
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    Re: Dimension of matrix spanning set

    and furthermore...oh snap! merry christmas!
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