Images of equivalent linear maps

Question

If we have linear maps $T_1, T_2 :U \rightarrow V$ with matrices $A$ and $B$ respectively and $A$ and $B$ are equivalent matrices, then what can we say about the images $\text{Im} (T_1) $ and $\text{Im}(T_2)$. It seems to me they are basically the same so we can say that they are isomorphic subspaces of $V$ but I can’t see how to write down this isomorphism.

Answer 1

I don't know exactly what happens if $T_1 \neq T_2$, but for the same $T$ you can construct with 2 different bases of $U$ 2 equivalent matrices whose Image are different. For example $T \in \mathit{L} (P_{1} (\mathbb{R}) ); T(p(x))=p´(x)$. Let $\beta=\{1,x\}$ and $\beta´ \{1+x,1-x\}$. Since \begin{equation} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}^{-1} = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}. \end{equation} And because there $\exists Q \in M_2 (\mathbb{R}), Q= [I]_{\beta´}^{\beta} \backepsilon [T]_{\beta´}=Q^{-1} [T]_{\beta} Q$, \begin{align} 1+x &= 1\cdot 1 + 1\cdot x \\ 1-x &= 1 \cdot 1 + (-1)\cdot x \end{align} \begin{equation} Q=\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \Rightarrow Q^{-1} = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix} \end{equation}. Now, \begin{align} T(1) &=0 \cdot 1 + 0 \cdot x \\ T(x) &=1 \cdot 1 + 0 \cdot x \end{align} So \begin{equation} [T]_{\beta}= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}. \end{equation} After some algebra you'll find that \begin{equation} [T]_{\beta´}= \begin{pmatrix} 1/2 & -1/2 \\ 1/2 & -1/2 \end{pmatrix} =\dfrac{1}{2} \begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix} \end{equation} Altough the range of both matrices is $1$; the vector that generates the Image of each transformation is not the same: $\begin{pmatrix} 1 \\ 0 \end{pmatrix} \neq \begin{pmatrix} 1 \\ 1 \end{pmatrix}$, hence their images aren't the same space.