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Let $D=A^TA$ where $A$ is an $m\times n$ matrix with $m>n$. Suppose that $D$ has a minimum eigenvalue $\mu>0$.

What can you say about the ratio $\frac{u^TDu}{u^Tu}$ for all vectors $u\neq 0$ ? In particular, how is this ratio related to $\mu$?

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I have been able to deduce that $D$ is invertible due to the min eigenvalue being strictly positive (no zero eigenvalues). $D$ is also symmetric due to being the product of a matrix and its transpose. I think $D$ being symmetric points towards it being a positive definite matrix. This would mean that the numerator is a positive value. The denominator would also be positive as a vector $v$ used in $v^Tv$ will always yield a positive answer.

Is this the limit to how far I can go with this? Can you deduce even more about the relation between $\mu$ and the ratio?

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The result you're looking for is Rayleigh's theorem. We have $$ \frac{u^TDu}{u^Tu} \geq \mu $$ and $\mu$ is attained when $u$ is the associated eigenvector of $D$.

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Since $D$ is a positive definite matrix, then an orthonormal basis $\{\mathbb{e_i}\}$ can be formed by its eigenvectors. Any vector $u$ can be decomposed as $u = \Sigma_i u_i\mathbb{e_i}$. If we use $\lambda_i$ denote the eigenvalues fo $D$, we have $\lambda_i \ge \mu$, then $u'Du = (\Sigma_i u_i^2\lambda_i)/\Sigma_i u_i^2 \ge (\mu\Sigma_iu_i^2)/(\Sigma_iu_i^2) \ge \mu$

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