Say $Q$ is a symmetric $n \times n$ matrix and $c \in R^n$ is non zero vector, and $\mu$ is a positive number.

Take the symmetric matrix $R = Q + \mu c c^T$. Let's denote the i'th eigenvector of a matrix $\lambda_i (A)$ and $\lambda_1 \leq \lambda_2 \leq ... \leq \lambda_n$

I want to show that for $n \geq 2$ then $\lambda_1(R) \leq \lambda_n(Q)$

I have managed to show that for the symetrix matrix Q, if it has an orthonormal basis of eigenvectors $w_1 , ... w_n$ and for R has $s_1 , ... ,s_n$

Then for some vector $v = a_1 w_1 + ... + a_n w_n$

The following holds: $\frac{v^TQv}{v^Tv} = \theta_1\lambda_1(Q) + ... + \theta_n \lambda_n(Q)$ for $\theta_i = \frac{a_i^2}{a_1^2 + ... + a_n^2}$

and $\lambda_n = max_{v \neq 0}\frac{v^TQv}{v^Tv}$ and $\lambda_1 = min_{v \neq 0}\frac{v^TQv}{v^Tv}$. Which respectively hold when $v = w_n$ and $v = w_1$

I managed to find that $\lambda_n(R) \geq \mu|c|^2 + \lambda_1(Q)$ Indeed: $\lambda_n(R) = \frac{s_n^TRs}{s_n^Ts} \geq \frac{w_1^TQw_1}{w_1^Tw_1} + \mu \frac{w_1^TQw_1}{w_1^Tw_1} \geq\lambda_1(Q) + \mu |c|^2$

Now I am stuck to show: $n \geq 2$ then $\lambda_1(R) \leq \lambda_n(Q)$

Any hints or insight would be greatly appreciated!


Observe that $$ \langle Qx,x \rangle \leq \langle Rx,x \rangle $$ (where $\langle \cdot, \cdot \rangle$ denotes the Euclidean scalar product). This means that you compare the quadratic forms associated with $Q$ and $R$.

This is tue since

\begin{align*} \langle Rx,x \rangle = \langle (Qx+ \mu c c^T) x,x \rangle &= \langle Qx,x \rangle + \mu \langle c c^T x ,x \rangle \\ &= \langle Qx,x \rangle + \mu (c c^T x)^T x \\ &= \langle Qx,x \rangle + \mu x^T c c^T x \\ &= \langle Qx,x \rangle + \mu (c^T x)^T c^T x \\ &= \langle Qx,x \rangle + \mu \| c^T x \|^2 \geq \langle Qx,x \rangle \end{align*}

Then apply Courant's minimax principle (https://en.wikipedia.org/wiki/Courant_minimax_principle) that express the eigenvalues of real symmetric matrices in terms of the associated quadratic forms.


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