# A problem on linear transformation .

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 $$W_1$$ .

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 $$M$$.