Finding the matrix for a projection map, $T^2=T$.

Question

Doing some revision, and was wondering if someone could help me work through this question.

You are given that $E:V\longrightarrow V$ is a projection map, so that $E^2=E$.

You are then asked to show that $E$ is a projection map if and only if there exists a basis of $V$ with respect to which $E$ can be written as a block matrix with $I$ (the identity matrix) in the top left corner, and the rest $0$.

I see the converse direction quite easily, if the matrix is of this block form, squaring it you clearly get the same matrix so therefore the $E^2=E$ is satisfied.

However, the forward direction I am struggling to see? I can't work out if I am meant to use the Jordan Normal Form, and if so, how, or if there is an easier way to do this. Any help, solutions, hints would be really appreciated. Thanks in advance.

Answer 1

Hint:

Let $E:V\to V$ a projection (I guess $V$ has finite dimensional). Let $\{e_1,...,e_n\}$ a basis of $V$. Then $\{E(e_1),...,E(e_n)\}$ are the eigenvectors associated to the eigenvalue $1$.

Answer 2

One way to see it is to consider the subspace $EV$. Take a basis $\{v_1,\ldots,v_k\}\cup\{w_1,\ldots,w_r\}$ of $V$ such that $\{v_1,\ldots,v_k\}$ is a basis of $EV$. Then you have $Ev_j=v_j$ for all $j=1,\ldots,k$, and $Ew_t=0$ for all $t=1,\ldots,r$ and so the matrix will have the desired form.

Another way is, as you say, to consider the Jordan form. If $E=SJS^{-1}$ then $J$ is a direct sum of Jordan blocks and $E^2=E$ implies $J^2=J$. This immediately gives you that the eigenvalues are $0$ and $1$, and by looking at the square of the Jordan blocks you conclude that all Jordan blocks have dimension 1.