Rank and Nullity of Transformation

Question

Let $B$ be any non-zero matrix of size $N \times N$ and let T be a transformation from $M_{N \times N} \text{ to } M_{N \times N}$, defined as : $T(A) = BA$.

What is the rank and nullity of $T$.

My answer: Find $\operatorname{Nullity}(T)$ by solving $BA = 0$. Then, $\operatorname{Rank}(T) = N^2 - \operatorname{Nullity}(T)$.

Can someone please confirm this? Also is there any direct formula for $\operatorname{Nullity}(T)$?

Answer 1

I assume you are working with $\mathbb{R}^{N}$.

By the rank null theorem you have

$rank(T)+null(T)=N^{2}$

Now to find null(T) you need to find the space of $N\times N$ matrices such that $BA=0_{N\times N}$. Define $C=BA$ then $c_{ij}=\langle b_{i,\cdot},a_{\cdot,j}\rangle$. So for each column $j$, finding $null(T)$ is equivalent to find $null(B)$ for linear space $B:N\to N$. Since $A$ is arbitrary then $null(T)=Nnull(B)$.

Answer 2

If $A=[a_1\ \dots\ a_N]$, then $A$ belongs to the null space of $T$ if and only if $Ba_i=0$, for $i=1,2,\dots,N$.

Thus the null space has dimension $Nd$, where $d$ is the nullity of $B$: can you see why and complete the proof?