Let $\rho(A)$ denote the spectral radius of a matrix $A$, defined as $\rho(A) = \underset{i}{\max} |\lambda_i|$, where $\lambda_i$ represents the $i$th eigenvalue of the matrix $A$.
In the course of solving a problem, I encountered the expression: $$ \rho(\alpha^2(AA^T)^2 - 2\alpha AA^T + I) $$ for a scalar $\alpha$ and matrix $A$. Upon experimenting with various matrices, I observed that linearity can be applied in this context, leading to the expression: $$ \rho(\alpha^2(AA^T)^2 - 2\alpha AA^T + I) = \alpha^2\rho(AA^T)^2 - 2\alpha\rho(AA^T) + 1 $$ I am struggling to comprehend the reason behind this phenomenon.
My attempted reasoning is as follows:
I understand that for a matrix $B$, we have $\rho(cB) = c\rho(B)$ for any scalar $c$. Additionally, I can demonstrate that $\rho(B+I) = \rho(B) + 1$. Assuming $\lambda$ is an eigenvalue of $B+I$:
$$
(B+I)v = \lambda v \implies Bv + v = \lambda v \implies Bv = (\lambda-1)v
$$
This implies that the eigenvalues of $B+I$ can be calculated by adding $1$ to the eigenvalues of $B$.
However, I am uncertain about how to proceed in proving $\rho(\alpha^2(AA^T)^2 - 2\alpha AA^T) = \alpha^2\rho(AA^T)^2 - 2\alpha\rho(AA^T)$.
I speculate that this phenomenon might be somehow related to the symmetry of $AA^T$, given that we are aware of $AA^T$ being a symmetric matrix. However, I am uncertain about the validity of this conjecture and how to proceed from this point.
Any assistance, hints, or insights would be greatly appreciated.
Your assertion is false. Let $\lambda_1\ge\lambda_2\ge\cdots\ge\lambda_n$ be the eigenvalues of $AA^T$, so that each $\lambda_i$ is nonnegative and $\rho(AA^T)=\lambda_1$. The eigenvalues of $D=\alpha^2 (AA^T)^2-2\alpha AA^T+I$ are then given by $$ \alpha^2\lambda_i^2-2\alpha \lambda_i+1 =(\alpha\lambda_i-1)^2. $$ It follows that $$ \rho(D)=\max_{1\le i\le n} (\alpha\lambda_i-1)^2 \quad\text{and}\quad \alpha^2\rho(AA^T)^2-2\alpha\rho(AA^T)+1 =(\alpha\lambda_1-1)^2. $$ These two expressions are guaranteed to be equal when the function $\lambda\mapsto(\alpha\lambda-1)^2$ is increasing. Since $\frac{d}{d\lambda}(\alpha\lambda-1)^2=2\alpha(\alpha\lambda-1)$, this means the two expressions are guaranteed to be equal when either $\alpha\le0$ or $\lambda_n\ge\frac{1}{\alpha}>0$. When this sufficient condition is not satisfied, counterexamples may occur. E.g. when $AA^T=A=\operatorname{diag}(1,0)$ and $\alpha=\frac12$, we have $$ \max\left\{(\alpha\cdot1-1)^2,(\alpha\cdot0-1)^2\right\}=1\ne\tfrac14=(\alpha\cdot1-1)^2. $$