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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.

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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

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

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  • $\begingroup$ Could you help explain what the Rayleigh quotient even is when it comes to these block matrices? $\endgroup$
    – mijucik
    Commented Mar 11, 2021 at 10:29
  • $\begingroup$ There is no difference! It is $x^{\top}Ax/x^{\top}x$ regardless of how $A$ is split into blocks. $\endgroup$
    – user140541
    Commented Mar 11, 2021 at 10:32
  • $\begingroup$ What are the dimensions of $x$ though? Are the entries of $x$ real? $\endgroup$
    – mijucik
    Commented Mar 11, 2021 at 10:34
  • $\begingroup$ $x$ should be compatible with $A$. $\text{dim}(x)=2n$ in your case. $x$ is a real vector when $A$ is real. $\endgroup$
    – user140541
    Commented Mar 11, 2021 at 10:36
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    $\begingroup$ 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. $\endgroup$
    – mijucik
    Commented Mar 11, 2021 at 10:41

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