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I have an asymmetric singular matrix $X \in \mathbb R^{n \times n}$ with real eigenvalues $\lambda_1 \geq \dots \geq \lambda_n$ and real linearly independent eigenvectors $q_1, \dots, q_n$, so that $X = Q \Lambda Q^{-1}$. I'll denote the eigenvalues of $X^T X$ by $\gamma_1 \geq \dots \geq \gamma_n$. Also let $P = I - \frac 1n \mathbb 1\mathbb 1^T$ (where $\mathbb 1$ is the vector of 1's of length $n$) and note that $Px$ is the projection of $x$ into the space orthogonal to $\mathbb 1$ (so $P$'s spectrum is the multiset $\{0,1,\dots,1\}$). I know a decent amount about the $\lambda_i$ and I want to bound (both upper and lower) the following two quantities in terms of the $\lambda_i$ if possible:

  1. $\gamma_1$, the largest eigenvalue of $X^T X$

  2. $\lambda_{max}(X^T P X) = \lambda_{max}\left((PX)^TPX\right)$

$X^T X$ is PSD so I know $0 \leq \gamma_1 \leq \textrm{tr}(X^T X)$ but this bound is too loose to be useful. It would be fantastic if there's a way to relate $\lambda_1$ and $\lambda_n$ to $\gamma_1$ in particular.

Both of these come down to being able to bound the largest eigenvalues of a matrix $A^TA$ in terms of the spectrum of $A$, so I could really use any results regarding that (like some version of Weyl's theorem for products). I know that problems like this are rarely answerable in total generality, but I'm hoping that there are some results that could apply (e.g. clever applications of Courant-Fischer). Thanks for any help.

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    If $X$ is singular, then how come we can have $n$ independent eigenvectors? – 2017-02-01
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    @Med - $X$ being singular just means at least one of the eigenvalues is 0. It does not in any way inhibit $X$ from having $n$ independent eigenvectors. – 2017-02-01
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    @alfalfa - could you explain what $1$ stands for when used as a column vector? This is not a notation I am familiar with. – 2017-02-01
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    @PaulSinclair it is just the vector of all 1's with the appropriate length – 2017-02-01

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