Show that there exists a nonzero column vector $x$ such that $A x = B x = 0$

Question

Given matrices $A, B \in \mathbb R^{n \times n}$ whose ranks satisfy the inequality $$r(A)+r(B) < n$$ show that there exists a vector $x \in \mathbb R^n \setminus \{0_n\}$ such that $A x = B x = 0_n$.

Since $r(A) + r (B) < n$ , hence $A, B$ both are singular matrices. Hence $Ax=0 $ and $Bx=0$ both have nonzero solutions. Now how to prove that intersection of the null space of $A$ and null space of $B$ is nontrivial.

Answer 1

Note that the dimension of the kernel of $A$ is $n - r(A)$, and a similar result holds for $B$. So there are $n - r(A)$ linearly independent vectors that form a basis of the kernel of $A$, and $n - r(B)$ linearly independent vectors that form a basis of the kernel of $B$. Note that $n - r(A) + n - r(B) > n$, so the two basis sets are not linearly independent, i.e. there exists a nonzero vector expressible in both bases. This vector is now our required $X$.

Answer 2

Let $d(A)=\dim(\ker A)$, $d(B)=\dim(\ker B)$. The rank-nullity formula implies that $\;d(A)+d(B)>n$.

Now, from the exact sequence $$0\longrightarrow\ker A\cap\ker B\longrightarrow \ker A\oplus\ker B\longrightarrow\ker A+\ker B\longrightarrow 0, $$ we deduce that $$\dim(\ker A\cap\ker B)=d(A)+d(B)-\dim(\ker A+\ker B)>n-n=0.$$ So the intersection has dimension at least $1$.