# Rayleigh Quotient with $2n \times 2n$ Matrices.

I need help understanding a problem.

For a bonus question in our problem set, we are told that $$\bf{A}$$ is a $$2n \times 2n$$ symmetric matrix of form $$\bf{A} = \begin{pmatrix} A_1 & A_3 \\ A_3 & A_2 \end{pmatrix}$$ where each entry is a $$n \times n$$ matrix, and that $$A_1$$ and $$A_2$$ are symmetric. We are then asked to prove that $$\lambda_\min (\bf{A}) \leq \min(\lambda_\min (A_1),\lambda_\min (A_2)).$$ We are asked to do this using the Rayleigh quotient $$\frac{x^TAx}{x^Tx}$$, which we've learned is bounded below and above by the smallest and largest eigenvalues of the matrix respectively, but is $$x$$ in this case supposed to be a vector of the form $$\begin{pmatrix} X_1 \\ X_2 \end{pmatrix}$$. If so, wouldn't this multiplication return a $$n \times n$$ matrix instead of a real value which can be bounded above and below by $$\lambda$$? It's all confusing me a bit, and I would appreciate it if someone could even just give me some definitions to work with, or some clarification on how the Rayleigh quotient could apply or work.

Let $$R(A,x)$$ denotes the Rayleigh quotient. Then $$\lambda_{\min}(A)=\min_x R(A,x) \le \min_{x:x_{n+1}=\ldots=x_{2n}=0} R(A,x)=\lambda_{\min}(A_1).$$ Similarly, $$\lambda_{\min}(A)=\min_x R(A,x) \le \min_{x:x_{1}=\ldots=x_{n}=0} R(A,x)=\lambda_{\min}(A_2).$$

Here

$$A=\left[ \begin{array}{ccc|ccc} a_{1,1} & \cdots & a_{1,n} & a_{1,n+1} & \cdots & a_{1,2n}\\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & \cdots & a_{n,n} & a_{n,n+1} & \cdots & a_{n,2n} \\ \hline a_{n+1,1} & \cdots & a_{n+1,n} & a_{n+1,n+1} & \cdots & a_{n+1,2n}\\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{2n,1} & \cdots & a_{2n,n} & a_{2n,n+1} & \cdots & a_{2n,2n} \\ \end{array}\right] \quad\text{and}\quad x=\left[ \begin{array}{c} x_{1} \\ \vdots \\ x_{n}\\ \hline x_{n+1} \\ \vdots \\ x_{2n} \\ \end{array}\right].$$

• Could you help explain what the Rayleigh quotient even is when it comes to these block matrices? Commented Mar 11, 2021 at 10:29
• There is no difference! It is $x^{\top}Ax/x^{\top}x$ regardless of how $A$ is split into blocks.
– user140541
Commented Mar 11, 2021 at 10:32
• What are the dimensions of $x$ though? Are the entries of $x$ real? Commented Mar 11, 2021 at 10:34
• $x$ should be compatible with $A$. $\text{dim}(x)=2n$ in your case. $x$ is a real vector when $A$ is real.
– user140541
Commented Mar 11, 2021 at 10:36
• I see I see. I think I had a misconception that the entries of the matrix were matrices itself, and not "numbers" in that sense. Now I realize the entries are just simply real values and the block matrix notation is just a notational thing and not a different mathematical structure. Commented Mar 11, 2021 at 10:41