0
$\begingroup$

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?

  • 0
    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

1 Answers 1

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}. $

  • 0
    yes,i mea$n$t (n-1)(n+$2$)\2 u may say it is a silly mistake:P2013-06-14