# How to prove $BA=0 \implies$ nullity $B \geq$ rank $A.$

I am stuck on the following problem and do not know how to progress:

Let $A$ and $B$ be $n \times n$ real matrices. Then how can I prove that $BA=0 \implies$ nullity $B \geq$ rank $A.$

Can someone point me in the right direction? Thanks in advance for your time.

• – Diego Silvera Mar 26 '13 at 3:59

Hint: This says that $\text{im}(A)\subseteq\ker B$--what's true about vector space dimension and containment?
• Sir,can you clarify a bit more about why " $\text{Im}(A)\subseteq\ker B$" and the given problem are equivalent. – user53386 Mar 26 '13 at 4:08
• @user53386 Everything that is in the iamge of $A$ is in the kernel of $B$ (why)? This says that the image of $A$ is contained in the kernel of $B$. But, the rank of $A$ is the dimension of its image, and the nullity of $B$ is the dimension of its kernel. Since this image of $A$ is contained in the kernel of $B$ this gives us the desired inequality (why?) – Alex Youcis Mar 26 '13 at 4:24
$A$ is a $n \times n$ matrix.then it goes a vector space with $n$ dimension to an another vector space also about $B!$ but when $BA=0$ is means that every base from image of $A$ will be at kernel of $B$. so $B$ at least has all of basis of image $A$, so null $B$≥ rank $A$.
• $\Huge{+1}~~~~$ – Mikasa May 21 '13 at 19:12
Another way of viewing this is that everything in the range of $A$ is sent to $0$ and is therefore in the Kernel of $B$. What does that say about the dimensions of these spaces?