Matrices with eigenvalue of multiplicity $k$ are algebraic subset of special linear group. In the previous exercise, I have showed that the special linear group $SL_n$ is a closed subvariety of $Mat(n,K)$ where $K$ is an algebraically closed field with characteristic zero. Now, I have to prove that the $SL_n^k$, the set of matrices $A \in SL_n$ with an eigenvalue of at least multiplicity $k$, is an algebraic subset of $SL_n$. 
So far, I know that $\det(A-\lambda I)$ splits in linear terms and one term has multiplicity of at least $k$. However, the hint is to use the resultant of two polynomials but I have no idea how I'm able to use this.
 A: Two tricks combine to solve this problem. The first is to observe that over a field of characteristic zero, a polynomial $p(x)$ of degree $d$ having a root $a$ of multiplicity $r$ is exactly equivalent to all derivatives up to order $r-1$ having $a$ as a root and the $r$th derivative not having $\lambda$ as a root if $r<d$. The second is that two polynomials $p(x),q(x)$ have a common root iff the resultant $Res_x(p(x),q(x))$ is zero.
To apply this to our situation, we consider the ideal generated by $Res_\lambda(\chi(\lambda),\chi(\lambda)^{(a)})$ for $0 < a < k$ where $\chi(\lambda)$ is the characteristic polynomial in the variable $\lambda$. Since these resultants are polynomials in the coefficients of the (derivatives of the) characteristic polynomial, which are themselves polynomials in the entries of the matrix, we see that we've got a perfectly good ideal of the coordinate algebra of $Mat(n,k)$ and that $V$ of this ideal exactly corresponds to those matrices with a repeated eigenvalue of order $k$.
