Consider a $m\times n$ matrix $A$ of rank $n$, where $n \leq m$. Let $W = A^TA$. I want to find a lower positive bound on the eigenvalues of $W$.
Because $A$ is full rank, I know that the eigenvalues of $W$ must be all positive. However, can I find a relationship with respect to the $A$? Can I say that the $\min_i\{\lambda_i(W)\}=\min_i\{\lambda_i(A)^2\}$ if $A$ is square? Where $\lambda_i(M)$ denotes the set the eigenvalues of a matrix $M$.
Extra info: For control theorists, the $A$ matrix here is the observability matrix and $W$ is the observability Gramian.
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I don't know if it is useful, but my $A$ matrix has special structure: $$ A = \begin{bmatrix} B \\ BC^{p_1} \\ BC^{p_2} \\ \vdots \\ BC^{p_i} \end{bmatrix}$$ where $B$, $C$ are of proper dimensions and $p_i>\dots>p_2>p_1>0$ are positive integers. The only special characteristic of $C$ is that it is an $n \times n$ invertible matrix. In that case we have that: $$W = B^TB+C^TB^T+\dots+C^{pT}B^TBC^p$$ Then, $\lambda_{min}(W) \geq \lambda_{min}(B^TB)+\lambda_{min}(C^TB^T)+\dots+\lambda_{min}(C^{pT}B^TBC^p)$ because all matrices in the summation are symmetric. But this bound doesn't help me much because it may easily ended up being zero, even though I know there is a higher bound.