Find basis that make my map look like the identity

Question

Let $k$ be a field. I have the map $$\displaystyle T = \begin{bmatrix} 1&1\\ 0&1 \end{bmatrix}$$ from $k^2$ to $k^2$, where I am using the standard basis in both spaces. My lecture notes claim that by changing the basis of the target space, I can turn this map into the identity. I have an inuition of why this happens: if instead of the standard basis I take the basis $\{e_1,e_1+e_2\}$ then the matrix should be the identity. But I would like a mathematical proof of this. I am not very good with computing the change of basis. If someone could indicate the operations I should be doing to obtain this matrix, I would really appreciate that.

Answer 1

If you have a linear map $T : V \to W$ and a basis $e_1,...,e_k$ of $V$ and $g_1,...,g_l$ of $W$ then the matrix of $T$ in those bases is by definition a matrix whose column $i$ are the coordinates of $T(e_i)$ in $W$.

For our matrix to be identity we therefore require $V$ and $W$ to have the same dimension and that $w_i := T(e_i)$ forms a basis of $W$.

Answer 2

Your intuition is correct.

Suppose $[T]_\beta$ is the matrix representation of $T:\Bbb F^2\to\Bbb F^2$ in the standard basis $\beta=\{e_1,e_2\}$ such that $$[T]_\beta=\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}.$$ Then we have \begin{align} (*) \ \ \ T(e_1)&=e_1;\\ (**) \ \ \ T(e_2)&=e_1+e_2. \end{align} Now, let $[T]_\beta^\gamma$ be the matrix representation of $T:\Bbb F^2\to\Bbb F^2$ in the standard basis $\beta=\{e_1,e_2\}$ and $\gamma=\{f_1,f_2\}$ such that $$[T]_\beta^\gamma=\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}.$$ Then we know that \begin{align} (\circ) \ \ \ T(e_1)&=f_1;\\ (\circ\circ) \ \ \ T(e_2)&=f_2. \end{align} From $(*)$ and $(\circ)$ we obtain $\boxed{f_1=e_1}$. From $(**)$ and $(\circ\circ)$ we obtain $\boxed{f_2=e_1+e_2}$.