Why do similar matrices have the same rank? I have seen some proofs on the Internet, which make use of the transformation map. However, I couldn't understand the methods since what I learned about the transformation map is so superficial. Can you use a simple way to explain it? And for now, just restrict the similar matrices to square matrices.
 A: 
Theorem. Similar matrices have the same rank.

Proof.
Part (1): We will show that $\text{rank}(AB)=\text{rank}(B)$ if $A$ is invertible.
Take some $u \in \text{ker}(B)$. Then, $Bu=0$, so $(AB)u=A(Bu)=A0 = 0$. Thus, $u\in \text{ker}(AB)$, so  $\text{ker}(B) \subseteq \text{ker}(AB)$.
Conversely, for some $u \in \text{ker}(AB)$, $(AB)u=0$, so $A(Bu)=0$. Thus $Bu=A^{-1}0=0$, so $U \in \text{ker}(B)$. Therefore, $\text{ker}(AB)= \text{ker}(B)$, implies $\text{null}(AB)=\text{null}(B)$, implies $\text{rank}(AB)=\text{rank}(B)$, provided that $A$ is nonsingular.
Part (2): Show that: $\text{rank}(AB)=\text{rank}(A)$, if $B$ is invertible.
$$\text{rank}(AB) = \text{rank}((AB)^{t})=\text{rank}(B^{t}A^{t})=\text{rank}(A^{t})=\text{rank}(A),$$
since $B^{t}$is nonsingular.
Thus, similar matrices have the same rank. We have $\text{rank}(XB)=\text{rank}(B)$ and $\text{rank}(AX)=\text{rank}(A)$ and $X^{-1}AX=B$ implies $AX=XB$.  Therefore, $\text{rank}(A)=\text{rank}(B)$.
A: By definition, $A\sim B$ iff there exists an invertible matrix $C$ such that $A=CBC^{-1}$ (and hence $AC=CB$). If $v$ is in the image of $B$, say $v=Bw$, then $Cv=CBw=ACw$ is in the image of $A$. Thus $C$ maps $\operatorname{Im}(B)\to \operatorname{Im}(A)$. As $C$ is injective, we conclude $\operatorname{rank}(A)\ge \operatorname{rank}(B)$. By symmetry, also $\operatorname{rank}(B)\ge \operatorname{rank}(A)$.
A: Here is an outline of a proof. First show that

Theorem. If $U$, $V$, and $W$ are linear spaces and $T: U \to V$, $S: V \to W$ are linear maps then the following holds.

*

*$\ker T \subseteq \ker S \circ T$ and $T(\ker S \circ T) \subseteq \ker S$.

*$\dim\ker S \circ T = \dim\ker T + \dim T(\ker S \circ T)$

*$\dim\ker T \leq \dim\ker S \circ T \leq \dim\ker T + \dim\ker S$

Now suppose $U=V=W$ and let $S=\phi$ be an invertible operator. Conclude that $\dim\ker \phi \circ T = \dim \ker T$. Interchanging the role of $S$ and $T$, you can conclude that $\dim\ker S \circ \phi = \dim\ker S$. This literally means that composition of an invertible operator from left or right with another operator does not change the dimension of the kernel. From this result conclude that the range and kernel of similar operators have the same dimension.
