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Diagonalizable matrix used in polynomial form? watch

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    If anyone you help with the last part of c i'd be very grateful!
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    What does the first part of (c) tell you about the span

    <B^n, B^(n-1), ... , B^2, B, I>?

    (PS the Oxford mods papers are online and may be easier to link to directly rather than producing a scan)
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    The idea is to use the identity from the first part of c) to re-write \displaystyle \Sigma a_i B^i in a more helpful form.
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    Hmm.. I must be being rather slow, still not really getting it. The first part of c tells us that the span <B^n, B^(n-1), ... , B^2, B, I> can equal zero for certain values of a[i] not equal to zero, so B^i are dependent? Don't know if that's helpful. Not sure about rewriting the span using the identity from the first bit, sorry!

    Oh wait, if B^i are linearly Dependant then they span the 3x3 matrices so there exists a linear combination that creates the required matrix?
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    If B^2 = (\lambda_1+\lambda_2)B - \lambda_1 \lambda_2 I, then what can be said about the set \{I,B,B^2\}, with regards to linear independence/dependence?

    Also, note that we cannot have \{I,B,B^2,\ldots,B^n\} being a linearly independent set for n \geq 9 (irrespective of what B actually is), since the dimension of the space of 3 x 3 matrices is 9, and a linearly independent set cannot have more elements than the dimension of the space it is in.

    So getting back to the question - it's possible to write the LHS of the equation at the end of C entirely in terms of B and I, can you see why?
 
 
 
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Updated: March 31, 2011
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