1
$\begingroup$

If $V$ and $W$ are two finite dimensional vector spaces and $\operatorname T$ and $\operatorname S$ be two invertible linear maps from $V$ to $W$, where $\operatorname T$ and $\operatorname S$ have same matrix representation wrt different ordered bases of $V$ and $W$.

Show that there exists two invertible linear maps $\operatorname{P}\colon V\longrightarrow V$ and $\operatorname{Q}\colon W\longrightarrow W$ such that $\operatorname T = \operatorname{Q^{-1}}\operatorname{S}\operatorname{P}$.

Actually, I was trying out by considering identity maps from $V$ onto $V$ wrt ordered bases $V_1$ and $V_2$ and the same for $W_1$ and $W_2$ but couldn't advance more.

Obviously firstly, I have to reach $V_2$ from $V_1$, then $W_2$ from $V_2$ and then $W_1$ from $W_2$ to make the composition of the maps equal to $\operatorname T$, i.e. from $V_1$ to $W_1$ .

$\endgroup$
1
$\begingroup$

So, you have two bases $B$ and $B^\star$ of $V$ and two bases $C$ and $C^\star$ of $W$ and you are assuming that the matrix of $T$ with respect to the basis $B$ and $C$ is equal to the matrix of $S$ with respect to the basis $B^\star$ and $C^\star$. Let $M$ be this matrix. Let $P\colon V\longrightarrow V$ be the linear endomorphism of $V$ which maps the $k$th vector of $B$ into the $k$th vector of $B^\star$. And let $Q\colon W\longrightarrow W$ be the linear endomorphism of $W$ which maps the $k$th vector of $C$ into the $k$th vector of $C^\star$. Then $T=Q^{-1}SP$ since the matrix of both linear maps with respect to the bases $B^\star$ and $C^\star$ are equal to $M$.

$\endgroup$
  • $\begingroup$ @ Jose Carlos Santos, Sir, can you please edit the text (as I am just new in learning Latex) ??? $\endgroup$ – Rabi Kumar Chakraborty Jul 28 at 18:57
  • $\begingroup$ I have done it. $\endgroup$ – José Carlos Santos Jul 28 at 19:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.