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Let $M_{n \times n}$ be the set of all $n\times n$ symmetric matrices such that the characteristic polynomial of each $A\in M_{n\times n}$ is of the form

$$t^n+t^{n−2}+a_{n−3}t^{n−3}+⋯+a_1t+a_0.$$

Then the dimension of $M_{n\times n}$ over $\mathbb{R}$ is

a) $(n−1)n/2$

b) $(n−2)n/2$

c) $(n−1)(n+2)/2$

d) $(n−1)^2/2$

For general symmetric matrices the dimension will be $n(n+1)/2$. What will it be here?

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    The question does not make a lot of sense, really. What is the conneection between the form of the characteristic polynomial and the dimension of the space of symmetric matrices?2012-11-17
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    Mariano, I think the point may be that the $t^{n-1}$ term is missing, so the trace is zero, etc.2012-11-17
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    kaikai, a lot of editing has been done to improve the formatting of your question. Please check to see whether all is well with the current form.2012-11-17
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    I'm a little worried about the coefficient of $t^{n-2}$ being a nonzero constant --- I'm not sure I believe such a set of matrices can be closed under addition.2012-11-17
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    @kaikai I've edited your title to be a bit more informative. If you prefer your old title then please rollback the edit.2012-11-17
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    @GerryMyerson, if $n=2$, then surely not :-)2012-11-17
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    @GerryMyerson It won't even be closed under scalar multiplication. The coefficient of $t^{n-2}$ gets scaled by $c^2$.2012-11-17
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    Well, it looks like we're agreed: the question is busted. kaikai, really, check to see whether this is what you wanted to ask.2012-11-17
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    If you add a coefficient $a_{n-2}$ in front of $t^{n-2}$, that should take care of things.2012-11-17
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    The question _does_ make sense as it is: even is the condition does not define a linear subspace, it does define an algebraic variety, which as such has a dimension. It is very doubtfull though that this was intended.2013-06-13

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