Engenvalues of $I_{T}+AA^{\prime }$

Question

Suppose $I_{T}$ is a $T\times T$ identity matrix, and $A$ is a $T\times m$ matrix where $m

Answer 1

Let $A = U\Sigma V^T$ be the singular value decomposition of $A$. Then $I_T + AA^T = U(I_T + \Sigma\Sigma^T)U^T$ is the eigendecomposition of $A$. The diagonal matrix of eigenvalues has the block form $$ I_T + \Sigma\Sigma^T = \begin{bmatrix} \begin{array}{ccc} 1 + \sigma_1^2 & &\\ & \ddots &\\ & & 1 + \sigma_r^2 \end{array} & O_{r\times(T-r)}\\ O_{(T-r)\times r} & I_{T-r} \end{bmatrix}, $$ where $r \leq m$ is the rank of $A$. Thus, the smallest eigenvalue of $A$ is $1$ which has algebraic multiplicity $T - r$ and the largest eigenvalue is $1 + \sigma_1^2$. Using matrix norm inequalities you can obtain the following upper bound on the maximum eigenvalue: $$ \lambda_\max = 1 + \sigma_1^2 \leq 1+\sqrt{\|I_T + AA^T\|_F^2 -2\|A\|_F^2 - T} $$ Naturally, many other such upper bounds are possible depending on what norm you choose.