Finding an orthonormal basis for a matrix If we have a subspace W of $\mathbb{R}^2$ spanned by $(3,4)$. Using the standard inner product, let E be the orthogonal projection of $\mathbb{R}^2$ onto W. Find an orthonormal basis in which E is represnted by the matrix:
$\begin{bmatrix}
    1      & 0  \\
    0      & 0 
\end{bmatrix}$
Now I figured out that $E(x_1,x_2) = \frac{1}{25}(9x_1+12x_2,12x_2+16x_2)$ and that the matrix for $E$ in the standard ordered basis is 
$\frac{1}{25}\begin{bmatrix}
    9      & 12  \\
   12      & 16 
\end{bmatrix}$. From here how would I figure out what the orthonormal basis is? I know that since the eigenvalues of $E$ are $\lambda =0,1$ then the basis for the other representation should be the unit eigenvectors. Would this be correct? It doesn't make sense to me because the orthonormal basis where the matrix representation of $E$ is the latter is just $\{(\frac{3}{5},\frac{4}{5})\}$ which does not span all of $\mathbb{R}^2$, but taking the eigenvectors would span all of $\mathbb{R}^2$.
 A: I think you are confused about the definition of "matrix representation of $E$" with respect to a basis. We regard $E$ as a linear transformation and ask "What does it do to the basis vectors of a given basis?" For the standard basis, 
$$
E(1,0)=1/25(9,12)
$$
and 
$$
E(0,1)=1/25(12,16).
$$
These vectors (regarded as column vectors) are the columns of the matrix representation of $E$ with respect to the standard basis for $\mathbb{R}^2$:
$$1/25
\begin{pmatrix}
9&12\\
12&16
\end{pmatrix}.
$$
Consider what $E$ does to the eigenvenvectors, (still with respect to the standard basis)
$$
E(3/5,4/5)=(3/5,4/5)
$$
and 
$$
E(-4/5,3/5)=(0,0).
$$
If we form a new basis from these eigenvectors,
we have
$$
(3/5,4/5)=1(3/5,4/5)+0(-4/5,3/5),
$$
so that $(3/5,4/5)$ is represented by $(1,0)$ in the new basis.
Similarly, 
$$(-4/5,3/5)=0(3/5,4/5)+1(-4/5,3/5),$$ so that it is represented by 
$(0,1)$ in the new basis.
Then, with respect to the new basis $E(1,0)=(1,0)$ and $E(0,1)=(0,0)$. The matrix that represents $E$ with respect to the new basis is then the matrix with these vectors as columns:
$$
\begin{pmatrix}
1&0\\
0&0
\end{pmatrix}.
$$
