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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.

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Hint: This says that $\text{im}(A)\subseteq\ker B$--what's true about vector space dimension and containment?

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  • $\begingroup$ Sir,can you clarify a bit more about why " $\text{Im}(A)\subseteq\ker B$" and the given problem are equivalent. $\endgroup$ – user53386 Mar 26 '13 at 4:08
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    $\begingroup$ @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?) $\endgroup$ – Alex Youcis Mar 26 '13 at 4:24
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$A$ 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 $A$.

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  • $\begingroup$ $\Huge{+1}~~~~$ $\endgroup$ – Mikasa May 21 '13 at 19:12
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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?

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