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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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3  
Mariano, I think the point may be that the $t^{n-1}$ term is missing, so the trace is zero, etc. –  Gerry Myerson Nov 17 '12 at 4:17
3  
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. –  Gerry Myerson Nov 17 '12 at 4:21
2  
@GerryMyerson It won't even be closed under scalar multiplication. The coefficient of $t^{n-2}$ gets scaled by $c^2$. –  EuYu Nov 17 '12 at 4:38
1  
Well, it looks like we're agreed: the question is busted. kaikai, really, check to see whether this is what you wanted to ask. –  Gerry Myerson Nov 17 '12 at 4:40
1  
If you add a coefficient $a_{n-2}$ in front of $t^{n-2}$, that should take care of things. –  Cameron Buie Nov 17 '12 at 4:52
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1 Answer 1

Since the $t^{n-1}$ term is missing so trace must be zero. Thus, we may consider a linear transformation

$$ T:S \to \mathbb R\qquad\text{s.t }\; T(A)=\operatorname{trace}(A) $$ where $A$ belongs to $S$ (the set of all symmetrical $n$-ordered matrices). Now all we need to prove that $T$ is surjective, so that we may apply Sylvester's law $$ \dim(S)= \operatorname{nullity}(T) + \operatorname{rank}(T) $$

But $\operatorname{rank}(T)= 1$ because $T$ is surjective; $\dim(S)=\frac{(n+1)n}{2}$; so, $$ \operatorname{nullity}(T)=\frac{(n-1)(n+2)}{2}. $$

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$\frac{(n+1)n}2-1\neq\frac{(n-1)(n-2)}2$, maybe you meant $\frac{(n-1)(n+2)}2$. –  Marc van Leeuwen Jun 13 '13 at 7:20
    
yes,i meant (n-1)(n+2)\2 u may say it is a silly mistake:P –  Geeta Jun 14 '13 at 5:02
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