Prove that $Nullity(AB)=Nullity(B)$ and $Range(AB)=Range(B)$

linear algebralinear-transformationsmatrices

Let be the field $K=\mathbb{R}$ or $\mathbb{C}$. $A \in K^{m\times n}$ is a
matrix associated to a linear transformation $A: K^n\rightarrow
K^m$, where $m,n \in Z_{+}$. We define:

\begin{align*} \operatorname{nullity} (A)=\dim (\ker A)&= \dim\left(\left \{ x \in K^n : Ax=0\right \}\right), \left \{ x \in K^n : Ax=0\right \} \subset K^n \\ \operatorname{range} (A)=\dim (\operatorname{Im}(A))&= \dim\left \{ y \in K^m, \exists x \in K^n : y=Ax \right \},\left \{\forall y \in K^m, \exists x \in K^n : y=Ax \right \} \subset K^m \end{align*}

  1. If $A\in K^{n\times n}$ is an invertible matrix, prove that $AB$ has the same nulity and range than B, for any matrix $B \in K^{n\times m}$
  2. If $m=n$, which is the relation between the eigenvalues and the eigenvectors of the matrices $AB$ and $BA$.

What I did for 1 is this:

\begin{align*}
ABx=0 \iff& Bx=0 \text{ since, } A \text{ is invertible}\\ \iff& x\in \operatorname{ker}(B) \ \ \forall x \in \operatorname{ker}(B)\\ \\ \Rightarrow \operatorname{nullity}(AB)=&\dim(\operatorname{ker} (AB))=\dim(ker(B)) =\operatorname{nullity}(B)
\end{align*}

By other side we have that,

\begin{align*}
\dim(K^m)=&\operatorname{nullity}(AB)+\operatorname{range}(AB)\\ \dim(K^m)=&\operatorname{nullity}(B)+\operatorname{range}(B)
\end{align*}

And also,
\begin{align*}
\dim(AB)=\dim(B) \ \ \ \ \ &\ \ \ \ \ \ \operatorname{nullity}(AB)=\operatorname{nullity}(B)\\ \\ \Rightarrow \operatorname{range}(AB)&=\operatorname{range}(B)
\end{align*}

Am I correct? Or what am I doing wrong?

And for 2 I'm not sure about which is the relation between the eigenvalues and the eigenvectors of the matrices AB and BA.

I would really appreciate your help!

Best Answer

The arguments for (1) are correct (mostly: remove the $b$ in $bBx=0$, and replace $\dim(B)$ by $\dim(K^m)$.)

For (2), suppose $ABv=\lambda v$, ($v\ne0$). Then $$BA(Bv)=B(ABv)=\lambda Bv$$ so $Bv$ is an eigenvector of $BA$ with eigenvalue $\lambda$, unless $Bv=0$ (i.e., $\lambda=0$).
Conversely, if $BAw=\mu w$ then $Aw$ is an eigenvector of $AB$, unless $Aw=0$.

Conclusion: The eigenvalues of $AB$ are the same as those of $BA$, except possibly for the zero eigenvalues (can't say in that case).

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