Show that $J-A$ will have nullity at least $9$ Show that if $A$ is a symmetric $n\times n$ matrix with nullity at least $10$ then $J-A$ will have nullity at least $9$ where $J$ is the all $1$ matrix of order $n\times n$. Since $A$ has nullity at least 10 so $0$ is an eigen value of $A$ with multiplicity at least 10. Also eigen values of $J$ are $n$ with multiplicity $1$ and $0$ with multiplicity $n-1$. How can I use them to show that $J-A$ has nullity at least $9$. Please help.

By rank-nullity theorem $$ \mathrm{rank}(A) = n - \mathrm{nul}(A) \le n-10. $$ Then, since $\mathrm{rank}(A+B) \le \mathrm{rank}(A) + \mathrm{rank}(B)$, $$ \mathrm{rank}(J-A) \le \mathrm{rank}(J) + \mathrm{rank}(-A) \le 1 + n-10 = n-9. $$ Using rank-nullity theorem again we get $$ \mathrm{nul}(J-A) = n - \mathrm{rank}(A) \ge 9. $$