Question: Let $E,F \subset \mathbb{R^n}$

Note that $< . >$ defines the Inner product on $\mathbb{R^n}$

Let $(e\_1,....,e\_k)$ and $(f\_1,.....f\_l$) be Orthonormal bases of E and F respectively.

Consider the Matrix $A=(<e\_i,f\_j>)$ of size $k\*l$

Let $\lambda\_i$ be an Eigenvalue of $A^T A$

Prove that $\lambda\_i \in [0, 1] \; \; \forall i$ and that they do not depend on the choice of bases of $E$ and $F$ 

I tryed experimenting on a small dimension matrix but I'm not able to figure it out. Thoughts?

Answer: 
If $V$ is any inner product space and $W$ is a subspace, then there is a natural orthogonal projection map $P\_W:V \to W$. These projections have several important properties. Among these we can show that $P\_W$ has operator norm equal to $1$ and is positive-semidefinite.
(<http://en.wikipedia.org/wiki/Hilbert_space#Orthogonal_complements_and_projections>)

We will consider the maps $P\_E: \mathbb{R}^n \to E$ and $P\_F: \mathbb{R}^n \to F$, and in particular we will consider the restrictions of these maps to appropriate subspaces.

Specifically, notice that $A^t$ is the matrix for $P\_F|\_E:E \to F$ with respect to the bases $(e\_1,\dots,e\_k)$ and $(f\_1,\dots,f\_l)$. Also notice that $A$ is the matrix for $P\_E|\_F:F \to E$ with respect to the bases $(f\_1,\dots,f\_l)$ and $(e\_1,\dots,e\_k)$. This can be easily computed using the definition of the matrix for a linear map and the properties of these projections.

It follows that $A^tA$ is the matrix for the map $P=P\_F \circ P\_E: F \to F$ with respect to the basis $(f\_1, \dots, f\_l)$. Since $P\_E$ and $P\_F$ have operator norm equal to $1$, it follows that $P$ has operator norm less than or equal to 1. Thus, all eigenvalues have absolute value less than or equal to $1$. But $P$ is also a positive semidefinite operator on $F$ (because for any $v \in F$, we have that $\langle Pv,v \rangle = \langle P\_Ev, P\_Fv \rangle = \langle P\_Ev, v \rangle = ||P\_Ev||^2\geq 0$), and thus the eigenvalues of $P$ must be non-negative. Then it immediately follows that the eigenvalues of $P$ (and thus $A^tA$) must be in $[0,1]$.

These eigenvalues only depend on the maps $P\_E$ and $P\_F$. In other words these eigenvalues only depend on the subspaces $E$ and $F$, and not the choices of bases. Also notice that this could have been done in any inner product space. Nothing special about $\mathbb{R}^n$ or the dot product.
Write $M\_{E}$ as the matrix with the columns $e\_j$ and $M\_{F}$ as the matrix with the columns $f\_j$, then notice that $A = M\_E^T M\_F$. Notice we are searching for $\left\|A \right\| = \sqrt{\lambda\_{max}(A^TA)}$ and trying to show that it is less than $1$. 
Now $\left\|A\right\| \leq \left\|M\_E \right\| \left\|M\_F\right\|$ and since $M^T\_E M\_E = I\_k$, $M^T\_F M\_F = I\_l$ we have $\left\|M\_E \right\| = 1 , \left\|M\_F \right\| = 1$ and finally yield $\left\|A\right\| \leq 1 $.