Basis of a projection Show that if $E$ is a projection on a finite dimensional vector space, then there exists a basis $B$ such that the matrix $(e_{ij})$ of $E$ with respect to $B$ has the following special form: $e_{ij}= 0$ or $1$ for all $i$ and $j$, and $e_{ij}=0$ if $i\neq j$. 
I seem to be having some difficulty understand this question. I could use a little help in the right direction.
 A: Added. In the context of linear algebra, a projection is any function $E\colon\mathbf{V}\to\mathbf{V}$ that is a linear transformation, and is such that $E^2=E$ (that is, applying $E$ twice is the same as applying it once.
For example, the map $E\colon\mathbb{R}^3\to\mathbb{R}^3$ that maps $(a,b,c)$ to $(a,b,0)$ is a projection: it is linear (check it), and $E^2(a,b,c) = E(E(a,b,c)) = E(a,b,0) = (a,b,0) = E(a,b,c)$. What does $E$ do to the standard basis? It sends $(1,0,0)$ to itself; it send $(0,1,0)$ to itself; and it sends $(0,0,1)$ to $(0,0,0)$. If we write the matrix of $E$ relative to the standard basis, we obtain
$$\left(\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0
\end{array}\right),$$
which agrees with the proposition you are being asked to proved.
Here's another example: take the linear transformation $E\colon\mathbb{R}^2\to\mathbb{R}^2$ that sends $(a,b)$ to $(\frac{1}{2}(a+b),\frac{1}{2}(a+b))$. This is a linear transformation (check), and
$$\begin{align*}
E^2(a,b) &= E\left(\frac{1}{2}(a+b),\frac{1}{2}(a+b)\right)\\
&= \left(\frac{1}{4}(a+b)+\frac{1}{4}(a+b), \frac{1}{4}(a+b)+\frac{1}{4}(a+b)\right)\\
 &= E(a,b),\end{align*}$$
so $E$ is a projection. Now, relative to the standard basis, the matrix of $E$ is not of the form you want:
$$\left(\begin{array}{cc}
\frac{1}{2} & \frac{1}{2}\\
\frac{1}{2} & \frac{1}{2}
\end{array}\right).$$
However, notice that a vector $(x,y)$ is mapped to itself under $E$ if and only if $(x,y)=(\frac{1}{2}(x+y),\frac{1}{2}(x+y))$, if and only if $\frac{1}{2}x=\frac{1}{2}y$, if and only if $x=y$. So the vectors that are mapped to themselves are all multiples of $(1,1)$. If we take $(1,1)$ as one of the basis vectors that we will put on $B$, then the column corresponding to $(1,1)$ will be all $0$s with a $1$ in the diagonal: exactly the kind we need. 
The second vector of $B$ has to be a vector that is mapped either to itself or to $(0,0)$ (since we want the matrix of $E$ relative to $B$ to be all $0$s except perhaps for some $1$s in the diagonal; think about what that means in terms of what $E$ does to the elements of $B$). But if the second vector is mapped to itself, then it will have to be a multiple of $(1,1)$ (because every vector that is mapped to itself is a multiple of $(1,1)$), and then $B$ will not be linearly independent. That means that the second vector in $B$ has to be a vector that is mapped to $(0,0)$ by $E$. A vector $(x,y)$ is mapped to $(0,0)$ if and only if $x+y=0$, if and only if $x=-y$. So if we pick $(1,-1)$ to be the second vector of $B$, that is, if $B=\{(1,1),(1,-1)\}$, then $B$ is a basis for $\mathbb{R}^2$ (check it!) and the matrix of $E$ with respect to $B$ is:
$$\left(\begin{array}{cc}
1 & 0\\
0 & 0
\end{array}\right),$$
which is of the desired form.

Now, in general...

You know that if $E$ is a projection, then $E=E^2$. Let $\mathbf{W}=\mathrm{Im}(E)$. Then how does $E$ act on $\mathbf{W}$? That is, what is $E|_{\mathbf{W}}$ (the restriction of $E$ to $\mathbf{W}$)? So, if you take a basis for $\mathbf{W}$, what does $E$ do to that bases?
That will be our start for $B$. We then need to complete $B$ so that the rest of the matrix for $E$ relative to $B$ is as described. From the description, we are going to need every other element of the basis to satisfy either $E\mathbf{x}=\mathbf{x}$ or $E\mathbf{x}=\mathbf{0}$. The former can't work, because that would mean that $\mathbf{x}\in\mathbf{W}$, which cannot happen (we already have a basis for $\mathbf{W}$ "in" $B$). So we are going to need every other element of the basis to be in the kernel of $E$. (The kernel is another name for the nullspace, in case that is the name you know it by).
Hmmm... Is that even possible? We would need the vector space to equal $\mathrm{W}+\mathrm{ker}(E) = \mathrm{Im}(E) + \mathrm{ker}(E)$. Since $\mathrm{ker}(E)\cap\mathrm{Im}(E)=\{\mathbf{0}\}$ (prove it!) that would mean that the vector space is $\mathrm{Im}(E)\oplus \mathrm{ker}(E)$.
Is this the case? If so, we'll be done! Take a basis for $\mathrm{ker}(E)$, take a basis for $\mathrm{Im}(E)$, put them together to get $B$. If not, then we're going to have to start over.
