Isomorphisms and Linear Transformations Suppose that $B = S^{-1}AS$ for some $n \times n$ matrices $A$, $B$, and $S$.


*

*Show that if $x \in \ker(B)$ then $Sx \in \ker(A)$.


Proof: $B = S^{-1}AS$ implies that $SB = AS$ which implies that $SBx = ASx = 0$, that is $Sx \in \ker(A)$.


*Show that the linear transformation $T : \ker(B) \to ker(A), \, x \mapsto Sx$ is an isomorphism. 


I know how to prove part 1, but I am not sure what to do for part 2. 
 A: $T:Ker\ (B)\rightarrow Ker\ (A)$ is given by $T(x)=S(x)$. First of all by part 1, this is well defined.


*

*Check that it is linear.

*Let $x\in Ker\ (T)$. Then $T(x)=S(x)=0\implies x\in Ker (S)$. But $S$ is invertible $\implies x=0$. So $T$ is one-one.

*Let $x\in Ker\ (A)\implies A(x)=0$. So consider the vector $S^{-1}x$. Then $BS^{-1}(x)=S^{-1}A(x)=0\implies S^{-1}(x)\in Ker\ (B)$ and $T(S^{-1}(x))=S(S^{-1}(x))=x$. So $T$ is onto.
Hence $T$ is an isomorphism.
A: A similar relationship exists with the roles of $A$ and $B$ reversed, and with $S$ replaced with its own inverse:
$$A=SBS^{-1}=(S^{-1})^{-1}BS^{-1}$$
This means that the conclusion from point 1 can be applied to this situation, as well: $S^{-1}$ maps the kernel of $A$ into the kernel of $B.$
A: The proof of 1 is good. Now, the map $T$ is well defined and obviously linear as it's the multiplication by a matrix.
If $x\in\ker T$, then $Sx=0$, so $x=0$ because $S$ is invertible. Therefore $T$ is injective. Since $A$ and $B$ have the same rank, their null spaces (kernels, in your terminology) have the same dimension (rank-nullity theorem) and injectivity implies surjectivity.
Alternatively, show that $U\colon \ker A\to\ker B$ is well defined by $y\mapsto S^{-1}y$. Obviously, $T$ and $U$ are inverse of each other.
