For symmetric definite matrix $A$ and $B$, prove that $k(A+B)\le \max(k(A),k(B))$ where $k(X)$ is the condition number of $X$

Question

Here the condition number is defined as the maximum eigenvalue divided by the minimum eigenvalue. For example, if $A=\text{diag}(a_1,a_2),B=\text{diag}(b_1,b_2)$,then $k(A)=\max(\frac {a_1}{a_2},\frac{a_2}{a_1})$,$k(B)=\max(\frac{b_1}{b_2},\frac {b_2}{b_1})$,$k(A+B)=\max(\frac {a_1+b_1}{a_2+b_2},\frac{a_2+b_2}{a_1+b_1})$. In this case,$k(A+B)\le \max(k(A),k(B))$ can be proved easily. However, for general cases, how can we give a simple proof? Thank you!

Answer 1

For any $n\times n$ symmetric matrix $M$, let $\lambda_1(M)\leq \ldots\leq \lambda_n(M)$ be the eigenvalues of $M$. Then you have, as a special case of the Courant-Weyl inequalities, that \begin{equation} \lambda_n(A+B)=\max_{x\neq 0} \frac{x^T(A+B)x}{x^Tx}\leq \max_{x\neq 0} \frac{x^TAx}{x^Tx}+\max_{x\neq 0} \frac{x^TBx}{x^Tx}=\lambda_n(A)+\lambda_n(B). \end{equation} Similarly, you can show that \begin{equation} \lambda_1(A+B)\geq \lambda_1(A)+\lambda_1(B). \end{equation} Then \begin{equation} k(A+B)=\frac{\lambda_n(A+B)}{\lambda_1(A+B)}\leq \frac{\lambda_n(A)+\lambda_n(B)}{\lambda_1(A)+\lambda_1(B)}. \end{equation} Can you take it from here?