Prove that there exist bases $\alpha $ and $\beta$ for V such that $ [T]_{\alpha}^{\beta} $ is a diagonal matrix with each diagonal entry equal to either 0 or 1.
Originally i thought that T=I was the only solution to this i realize that is not the case now but i am still lost what if T forms an non-invertable matrix? i really dont understand how we can always know that this is true cause if the map is non-invertable isn't it not diagonalizable?
On
Suppose that $T$ has rank k, then you can construct a basis made of $n-k$ vectors in Ker(T) and of $k$ vectors orthogonal to them
$$\alpha=\{\alpha_1,...\alpha_k,\alpha_{k+1},...\alpha_n\}$$
then assume
$$\beta=\{\beta_1,...\beta_k,\beta_{k+1},...\beta_n\}$$
with $\beta_i=T(\alpha_i)$ thus $[T]_{\alpha}^{\beta}$ is a diagonal matrix with each diagonal entry equal to either 0 or 1
Let $T:V\rightarrow V$ be a linear map and $\alpha=\left\{v_1, \dots, v_k, u_{k+1}, \dots u_n\right\}$ a basis of $V$ such that $\left\{v_1, \dots, v_k\right\}$ is a basis of $\ker(T)$.
You can easily prove that $T(u_{k+1}), \dots , T(u_n)$ are linearly independent vectors. (Mimick the technique used to prove the rank-nullity theorem).
Define $\beta=\left\{w_1, \dots, w_k,T(u_{k+1}), \dots , T(u_n) \right\}$. Here $w_1, \dots, w_k$ are vectors extending the linearly independent $T(u_{k+1}), \dots , T(u_n)$ to a basis of $V$. Now you can easily check that $[T]_{\alpha}^{\beta}$ is of the required form.
Notice that this does not imply that $\det(T)=0$ or $1$. Indeed, $\det(T)$ is defined as $\det([T]_{\gamma}^{\gamma})$ for some basis $\gamma$ of $V$. So the matrix of $T$ is expressed w.r.t. the same basis in both domain and target.