Linear Algebra – Proving $\dim(\text{range}(S^{-1}TS)) = \dim(\text{range}(T))$ and $\operatorname{range}(S^{-1}TS) = \operatorname{range}(T)$

linear algebralinear-transformations

Prove or disprove: $\dim \left( \operatorname{range} \left( S^{-1} T S \right) \right) = \dim \left( \operatorname{range} \left( T \right) \right)$, where $S$ is invertible and $T, S \in L(V)$.

I intuitively think it is true, because $S$ is invertible and it doesn't change the column space and row space of $M(T)$, where $M(T)$ is the matrix of $T$. But I cannot give a strict proof.

Or if I can prove $\dim \left( \operatorname{null} \left( S^{-1} T S \right) \right) = \dim \left( \operatorname{null} \left( T \right) \right)$, I can say $\dim \left( \operatorname{range} \left( S^{-1} T S \right) \right) = \dim \left( \operatorname{range} \left( T \right) \right)$ because of Fundamental theorem of linear algebra. But I failed.

And what's more I wonder whether $\operatorname{range} \left( S^{-1} T S \right) = \operatorname{range} \left( T \right)$ or not.

Best Answer

  • First, we must prove that $\operatorname{range}\left(S^{-1}TS\right)=\operatorname{range}\left(S^{-1}T\right).$

Proof. Let $x\in\operatorname{range}\left(S^{-1}TS\right)$ be arbitrary, then $x=S^{-1}TS(v)$ for some $v\in V$. Since $x=S^{-1}TS(v)=S^{-1}T(S(v))$ for some $S(v)\in V$, then $x\in\operatorname{range}\left(S^{-1}T\right)$. Therefore, $\operatorname{range}\left(S^{-1}TS\right)\subseteq\operatorname{range}\left(S^{-1}T\right).$

Let $x\in\operatorname{range}\left(S^{-1}T\right)$ be arbitrary, then $x=S^{-1}T(v)$ for some $v\in V$. Since $x=S^{-1}T(v)=S^{-1}T(SS^{-1}(v))=S^{-1}TS(S^{-1}(v))$ for some $S^{-1}(v)\in V$, then $x\in\operatorname{range}\left(S^{-1}TS\right)$. Therefore, $\operatorname{range}\left(S^{-1}T\right)\subseteq\operatorname{range}\left(S^{-1}TS\right).$

Therefore, $\operatorname{range}\left(S^{-1}TS\right)=\operatorname{range}\left(S^{-1}T\right).$ In particular, $$\tag{*} \label{*} \dim\operatorname{range}\left(S^{-1}TS\right)=\dim\operatorname{range}\left(S^{-1}T\right).$$

  • Second, we must prove that $\operatorname{range}\left(S^{-1}T\right)\simeq\operatorname{range}\left(T\right).$

Proof. Let $\varphi:\operatorname{range}\left(T\right)\to\operatorname{range}\left(S^{-1}T\right)$ be a linear transformation defined by $\varphi(x)=S^{-1}x$.

Let $x,y\in\operatorname{range}\left(T\right)$. Note that $$\varphi(x)=\varphi(y)\implies S^{-1}x=S^{-1}y\implies SS^{-1}x=SS^{-1}y\implies x=y.$$ Therefore, $\varphi$ is injective.

Now, let $y\in\operatorname{range}\left(S^{-1}T\right)$ be arbitrary. Choose $x=Sy$. Thus, $x\in\operatorname{range}\left(T\right)$ and $$\varphi(x)=\varphi(Sy)=S^{-1}\left(Sy\right)=y.$$ Therefore, $\varphi$ is surjective.

Therefore, $\operatorname{range}\left(S^{-1}T\right)\simeq\operatorname{range}\left(T\right).$ It follows that $$\tag{**} \label{**}\dim\operatorname{range}\left(S^{-1}T\right)=\dim\operatorname{range}\left(T\right).$$

  • Finally, it follows from $(\ref{*})$ and $(\ref{**})$ that $$\boxed{\dim\operatorname{range}\left(S^{-1}TS\right)=\dim\operatorname{range}\left(T\right)}.$$
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