Showing a linear transformation is diagonalisable Let $f:V\to V$ be a linear map of a finite dimensional vector space $V$ with $f^2=f$. Show that $f$ is diagonalisable and determine the possible eigenvalues of $f$.

So we know $f^2=f$, or in other words $f(f-1)=0$. That is, $p(x) = x(x-1)$ is the characteristic polynomial of $f$, meaning possible eigenvalues of $f$ are $0$ and $1$. The minimal polynomial $m(x)$ divides $p(x)$ so we have three options $m(x) = x$, $m(x) = x-1$ or $m(x)=x(x-1)$. 
Now I don't really know how to link this to showing $f$ is diagonalisable, any tips?
 A: Do not miss the geometry involved! That is what is really interesting about this problem. Not the computations, which you actually do not need. If $f^2=f$ then $f$ is a projection operator. Do you see why? You are projecting into the image of $f$, which of course is a linear subspace. In fact this gives you already the whole answer without computations.
You know that $f$ is diagonalisable if and only if $V$ has a basis composed of eigenvectors of $f$. 
You must know at this point that $V$ can be decomposed as $$V=\mathrm{ker}(f) \oplus \mathrm{im}(f)$$
Any element of the kernel will have $f(v)=0$. Any element of the image will have $f(v)=v$. Pick a basis for the kernel and a basis for the image. Put the together and you will get a basis of eigenvectors of $f$. You also conclude that the only possible eigenvectors are zero and one.

Without geometry (This is the most painful but must basic path).
Suppose you have an eigenvector $v \neq 0$ associated to eigenvalue $\lambda$, we have $f(v)=\lambda v$. Since $f^2=f$ then 
$$\lambda v=f(v)=f^2(v)= f(f(v)) =f(\lambda v) = \lambda^2 v$$
and since $v$ is non-zero you deduce that either $\lambda=1$ or $\lambda=0$. 
This implies that one eigenspace is the kernel, which, by definition, is the subspace of vectors satisfying $f(v)=0$; and that the other eigenspace is the image of $f$ (for any element $v$ that can be written as $f(u)=v$ you have $v=f(u)=f^2(u)=f(f(u))=f(v)$, so every element if the image is an eigenvector associated to $\lambda=1$). 
Choose now a maximal independent set $\alpha_1, ...,\alpha_r$ in the kernel and a maximal independent set $\beta_1,...,\beta_s$ in the column space of $f$ (that is, n its image). The rank-nullity theorem which you should know (follow the link to see a proof if you don't) tells you that sum of $r$ and $s$ is the dimension of $V$. The rank-nullity already implies then, that we can pick a basis of the image and a basis of the kernel, and that together they form a basis of $V$. By the criterion saying that $f$ is diagonalizable if and only if there is a basis of eigenvectors we may conclude the result.
If you want an explicit construction do the following. Define a diagonal matrix $D$ such that its diagonal is the vector $(0,...,0,1,...,1)$ where $0$ appears $r$ times and $s$ appears $1$ times. Define the matrix $P=(\alpha_1|...|\alpha_r|...|\beta_1|...|\beta_s)$ where the vertical means that each column is the given vector. Now you only need to convince your self that $f=PDP^{-1}$: you can do this by checking that both sides are the same applied to some basis. The most reasonable thing to do is to apply it to $\alpha_1,...,\alpha_r,\beta_1,...,\beta_s$. . In fact you only need to check that $PDP^{-1}\alpha_i=0$ for all $\alpha_i$ and $PDP^{-1}\beta_i=\beta_i$ for all $\beta_j$. 
A: Claim: From $f^2=f$ it follows (a) that  $f$ is the identity on the subspace ${\rm im}(f)\subset V$ and (b) that ${\rm im}(f)\oplus{\rm ker}(f)=V$.
Proof. (a) If $y=f(x)$ then $f(y)=f^2(x)=f(x)=y\ .\ $ (b) Consider an $y\in {\rm im}(f)\cap{\rm ker}(f)$. Then $y=f(x)=f\bigl(f(x)\bigr)=f(y)=0$. As ${\rm dim}\bigl({\rm im}(f)\bigr)+{\rm dim}\bigl({\rm ker}(f)\bigr)={\rm dim}(V)$ the claim follows.$\quad\square$
Choose a basis $(f_1,\ldots, f_r)$ of ${\rm im}(f)$ and a basis $(f_{r+1},\ldots, f_n)$ of ${\rm ker}(f)$. Then $(f_1,\ldots, f_n)$ is a basis of $V$, and with respect to this basis $f$ has the matrix ${\rm diag}(1^r, 0^{n-r})$.
