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 $9$.

• Hint: the null space of $J - A$ contains the intersection of the null spaces of $J$ and of $A$ – Christopher Jun 14 '18 at 8:42
• @user73985;The null space of $J$ has dimension $n-1$ and that of $A$ is $10$ so how can it be $9$ for $J-A$ – Learnmore Jun 14 '18 at 8:50
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 $$\mathrm{nul}(J-A) = n - \mathrm{rank}(A) \ge 9.$$