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I'm studying inequalities involving positive definite symmetric matrices and their principal submatrices. Let $M$ be a positive definite symmetric matrix, partitioned as follows: $$ M = \begin{pmatrix} M_{11} & M_{12} \\ M_{21} & M_{22} \end{pmatrix} $$ where $M_{11}$ is the leading principal submatrix of $M$, and $M_{12} = M_{21}'$ represent the off-diagonal blocks. As $M$ is positive definite we know that $x'Mx > 0$ for all non-zero vectors $x$. Moreover, let's write $x = (x_1, x_2)$, then we also know that $x_1' M_{11}x_1 > 0$ and $x_2'M_{22}x_2 > 0$ for all non-zero vectors $x_1$ and $x_2$ of appropriate length (see this post).

Specifically, I’m interested in understanding when the inequality $c \cdot x_1' M_{11} x_1 \leq x' M x $ might hold for some constant $c > 0$.

Clearly, when the off-diagonals blocks are zero, this inequality holds for $c = 1$. However, when these are non-zero, this does not hold for $c = 1$. Here’s an example that demonstrates this.

Define $M$ as $$ M = \begin{pmatrix} 5 & 2 \\ 2 & 1 \end{pmatrix}, $$ and let $x = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$. Then, $$ x' M x = \begin{pmatrix} 2 & -1 \end{pmatrix} \begin{pmatrix} 5 & 2 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = 13. $$

Now consider the leading principal submatrix $M_{11} = \begin{pmatrix} 5 \end{pmatrix}$ and vector $x_1 = \begin{pmatrix} 2 \\ 0 \end{pmatrix}$: $$ x_1' M_{11} x_1 = 2 \cdot 5 \cdot 2 = 20. $$

In this case, we find that $x' M x = 13$ while $x_1' M_{11} x_1 = 20$, so the inequality $x' M x \geq x_1' M_{11} x_1$ does not hold.

So my question is: For positive definite symmetric matrices $M$, does there exist a constant $c > 0$ such that the inequality $c \cdot x_1' M_{11} x_1 \leq x' M x$ always holds?

Any insights or references on this question would be highly appreciated!

Stan
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1 Answers1

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No, such a $c$ does not exist.

To demonstrate that there does not exist a constant $c$ such that $c \cdot x_1' M_{11} x_1 < x' M x$ holds universally for any positive definite matrix $M$ and any vector $x$, we consider the following construction.

Define the matrix $$ M = \begin{pmatrix} 1 & 1 - \epsilon \\ 1 - \epsilon & 1 \end{pmatrix}, $$ where $\epsilon$ is a small positive parameter that we can adjust. Note that $M$ is symmetric and positive definite.

Let $x = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$ and consider the quadratic forms $x' M x$ and $x_1' M_{11} x_1$, where $M_{11} = (1)$ is the leading principal submatrix of $M$, corresponding to the upper-left $1 \times 1$ block.

  1. Calculate $x' M x$: $$ x' M x = \begin{pmatrix} 1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 1 - \epsilon \\ 1 - \epsilon & 1 \end{pmatrix} \begin{pmatrix} 1 \\ -1 \end{pmatrix}. $$ Expanding this product, we find: $$ x' M x = 1 \cdot 1 + (-1)(1 - \epsilon) + (1 - \epsilon)(-1) + (-1)(-1) = 2 \epsilon. $$

  2. Calculate $x_1' M_{11} x_1$: Since $M_{11} = (1)$, we have $$ x_1' M_{11} x_1 = 1 \cdot 1 \cdot 1 = 1. $$

  3. Comparison and Limiting Behavior: Now, $x' M x = 2 \epsilon$ can be made arbitrarily close to zero by taking $\epsilon$ very small, while $x_1' M_{11} x_1 = 1$ remains fixed. As a result, no constant $c$ exists such that $$ c \cdot x_1' M_{11} x_1 < x' M x $$ holds for all positive definite matrices $M$ and all choices of $x$, because $x' M x$ can be made arbitrarily close to $x_1' M_{11} x_1$ by choosing a sufficiently small $\epsilon$.

This example illustrates that the inequality cannot hold universally with a fixed constant $c$.

Stan
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