Suppose $A$ is a real $n\times n$ matrix of rank $r$. Let $V$ be the vector space of all real $n\times n$ matrices $X$ such that $AX=O$. What is the dimension of $V$?
My working: We need matrices $X$ such that $AX=O$ should form the null space of $T$. Hence from rank nullity theorem Nullity $(T)=n^2-r$.
But the right answer is $n^2-nr$.
Please correct me and help me to solve this problem.
Since ${\rm rank}A=r$ and the matrix $X$ has $n$ columns, the image of map $T(X)=AX$ has the dimension of $nr$. Applying the rank nullity theorem, we have $\dim V=n^2-nr$.
(If you are still not sure, you can use this approach: $X$ has $n$ columns, so instead of working with $V$, you can work with each column of $X$. For each column of $X$ the null space is generated by $n-r$ vectors $v_1,...,v_r$ (since ${\rm rank}A=r$). Now write-everything-down for all the columns of $X$ you will get the same result as above)
