<s>[INST] Linear algebra inequality (1) Note that if $\| \mathbf{A} \mathbf{x} \|_{2} = 1$, then 
$$\| \mathbf{A}^{T} \mathbf{A} \mathbf{x} \|_{2} \cdot \| \mathbf{x} \|_{2} \ge \langle \mathbf{A}^{T} \mathbf{A} \mathbf{x}, \mathbf{x} \rangle = \| \mathbf{A} \mathbf{x} \|_{2}^{2} = 1$$
and thus we have $\| \mathbf{A}^{T} \mathbf{A} \mathbf{x} \|_{2} \ge \frac{1}{\| \mathbf{x} \|_{2}}$.
(2) By (1), if $\| \mathbf{A} \mathbf{x}_{1} \|_{2} = \cdots = \| \mathbf{A} \mathbf{x}_{r} \|_{2} = 1$, we have
$$\| \mathbf{A}^{T} \mathbf{A} \mathbf{x}_{1} \|_{2}^{2} + \cdots + \| \mathbf{A}^{T} \mathbf{A} \mathbf{x}_{r} \|_{2}^{2} \ge \frac{1}{\| \mathbf{x}_{1} \|_{2}^{2}}+\cdots+\frac{1}{\| \mathbf{x}_{r} \|_{2}^{2}}.$$
Here is my question:
If $\{ \mathbf{A}\mathbf{x}_{1}, \cdots, \mathbf{A}\mathbf{x}_{r} \}$ is orthonormal, can we obtain a better inequality than (2)?
Thanks.
 [/INST] Consider $A=I$ and $x_i=e_i$ the $i$-th vector of the canonical basis. Then  the set $\{ \mathbf{A}\mathbf{x}_{1}, \cdots, \mathbf{A}\mathbf{x}_{r} \}$ is orthonormal, but :
$$\| \mathbf{A}^{T} \mathbf{A} \mathbf{x}_{1} \|_{2}^{2} + \cdots + \| \mathbf{A}^{T} \mathbf{A} \mathbf{x}_{r} \|_{2}^{2} =r= \frac{1}{\| \mathbf{x}_{1} \|_{2}^{2}}+\cdots+\frac{1}{\| \mathbf{x}_{r} \|_{2}^{2}}.$$
Hence the inequality cannot be "better" even if we take orthogonality into account.
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