91Ó°ÊÓ

$$\begin{gathered} \mathrm{If}{\mathrm{S}}^{-1}\mathrm{AS}=\mathrm{B},\mathrm{then}\mathrm{AS}=\mathrm{SB}.\mathrm{Now},\mathrm{it}\mathrm{applies} \\ \mathrm{x}∈\ker \mathrm{B}⇔\mathrm{Bx}=0⇔\mathrm{ASx}=\mathrm{SBx}=\mathrm{SBx}=\mathrm{S}\left(0\right)=0 \end{gathered}$$

Let :$\mathrm{T}:\ker \mathrm{B}→\ker \mathrm{A},\mathrm{T}\left(\mathrm{x}\right)=\mathrm{Sx}.$

As a restriction of S, we know that T is linear. To prove it's an isomorphism, we need to prove it's both a monomorphism and an epimorphism.

It applies

$x∈\ker \mathrm{T}⇔\mathrm{Tx}=0⇔\mathrm{Sx}=0⇔\mathrm{x}=0$.

The last equivalence applies due to S being invertible, therefore an isomorphism.T Thus, is a monomorphism

Therefore,T is an isomorphism.

If an invertible nxn matrix is multiplied by another nxn matrix of rank $\mathrm{k}≤\mathrm{n}$, the rank of the product will be k. So we have $\mathrm{A}=\mathrm{rank}\left({\mathrm{S}}^{-1}\mathrm{AS}\right)=\mathrm{rankB}$. Then, it directly applies.

nullity A = nullity B

Here, the final result of part(a), part(b), part(c) is$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{x}∈\ker \mathrm{B}⇔\mathrm{Bx}=0⇔\mathrm{ASx}=\mathrm{SBx}=\mathrm{s}\left(0\right)=0 \\ \left(\mathrm{b}\right)\mathrm{x}∈\ker \mathrm{T}⇔\mathrm{Tx}=0⇔\mathrm{Sx}=0⇔\mathrm{x}=0\mathrm{T}\mathrm{is}\mathrm{an}\mathrm{isomorphism}. \\ \left(\mathrm{c}\right)\mathrm{it}\mathrm{directly}\mathrm{applies}\mathrm{nullity}\mathrm{A}=\mathrm{nullity}\mathrm{B} \end{gathered}$$