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Brian Borchers
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Here's a restatement of the problem for those who don't want to find the paper referenced in the question.

We're interested in a symmetric and positive definitive matrix $\Sigma^{-1}$$\Sigma$ that has been orthogonally diagonalized as

$ \Sigma=Q^{T}\mbox{diag}(\theta_{1}^{2},...,\theta_{n}^{2})Q $

where $\theta_{1}^{2} \geq \theta_{2}^{2} \geq ... \geq \theta_{n}^{2} > 0$, and we let $\theta_{\max}^{2}=\theta_{1}^{2}$ and $\theta_{\min}^{2}=\theta_{n}^{2}$.

Also, let

$\Theta^{2}=\mbox{diag}(\theta_{1}^{2},...,\theta_{n}^{2})$.

We want to show for any nonzero vector $\alpha$,

$ \frac{\alpha^{T}\Sigma^{-1}\alpha}{\sqrt{\alpha^{T}\alpha}\sqrt{\alpha^{T}\Sigma^{-2}\alpha}} \geq \frac{\theta_{\max}\theta_{\min}}{(\theta_{\max}^{2}+\theta_{\min}^{2})/2} $

Note that you can assume without loss of generality that $\alpha$ is of length 1. (just scale the length of alpha out of everything on the left hand side of the inequaility.)

Also, by using the substitution $x=Q\alpha$, you can reduce this to a problem about the digonal matrix $\Theta$, and then reduce the matrix-vector products to sums. We then want to show for all vectors $x$ of length 1,

$ \frac{\sum_{i=1}^{n} x_{i}^{2}\theta_{i}^{-2}} {\sqrt{\sum_{i=1}^{n}x_{i}^{2}\theta_{i}^{-4}}} \geq \frac{\theta_{\max}\theta_{\min}}{(\theta_{\max}^{2}+\theta_{\min}^{2})/2} $

Unfortunately, I don't see any easy way to proceed from here.

Here's a restatement of the problem for those who don't want to find the paper referenced in the question.

We're interested in a symmetric and positive definitive matrix $\Sigma^{-1}$ that has been orthogonally diagonalized as

$ \Sigma=Q^{T}\mbox{diag}(\theta_{1}^{2},...,\theta_{n}^{2})Q $

where $\theta_{1}^{2} \geq \theta_{2}^{2} \geq ... \geq \theta_{n}^{2} > 0$, and we let $\theta_{\max}^{2}=\theta_{1}^{2}$ and $\theta_{\min}^{2}=\theta_{n}^{2}$.

Also, let

$\Theta^{2}=\mbox{diag}(\theta_{1}^{2},...,\theta_{n}^{2})$.

We want to show for any nonzero vector $\alpha$,

$ \frac{\alpha^{T}\Sigma^{-1}\alpha}{\sqrt{\alpha^{T}\alpha}\sqrt{\alpha^{T}\Sigma^{-2}\alpha}} \geq \frac{\theta_{\max}\theta_{\min}}{(\theta_{\max}^{2}+\theta_{\min}^{2})/2} $

Note that you can assume without loss of generality that $\alpha$ is of length 1. (just scale the length of alpha out of everything on the left hand side of the inequaility.)

Also, by using the substitution $x=Q\alpha$, you can reduce this to a problem about the digonal matrix $\Theta$, and then reduce the matrix-vector products to sums. We then want to show for all vectors $x$ of length 1,

$ \frac{\sum_{i=1}^{n} x_{i}^{2}\theta_{i}^{-2}} {\sqrt{\sum_{i=1}^{n}x_{i}^{2}\theta_{i}^{-4}}} \geq \frac{\theta_{\max}\theta_{\min}}{(\theta_{\max}^{2}+\theta_{\min}^{2})/2} $

Unfortunately, I don't see any easy way to proceed from here.

Here's a restatement of the problem for those who don't want to find the paper referenced in the question.

We're interested in a symmetric and positive definitive matrix $\Sigma$ that has been orthogonally diagonalized as

$ \Sigma=Q^{T}\mbox{diag}(\theta_{1}^{2},...,\theta_{n}^{2})Q $

where $\theta_{1}^{2} \geq \theta_{2}^{2} \geq ... \geq \theta_{n}^{2} > 0$, and we let $\theta_{\max}^{2}=\theta_{1}^{2}$ and $\theta_{\min}^{2}=\theta_{n}^{2}$.

Also, let

$\Theta^{2}=\mbox{diag}(\theta_{1}^{2},...,\theta_{n}^{2})$.

We want to show for any nonzero vector $\alpha$,

$ \frac{\alpha^{T}\Sigma^{-1}\alpha}{\sqrt{\alpha^{T}\alpha}\sqrt{\alpha^{T}\Sigma^{-2}\alpha}} \geq \frac{\theta_{\max}\theta_{\min}}{(\theta_{\max}^{2}+\theta_{\min}^{2})/2} $

Note that you can assume without loss of generality that $\alpha$ is of length 1. (just scale the length of alpha out of everything on the left hand side of the inequaility.)

Also, by using the substitution $x=Q\alpha$, you can reduce this to a problem about the digonal matrix $\Theta$, and then reduce the matrix-vector products to sums. We then want to show for all vectors $x$ of length 1,

$ \frac{\sum_{i=1}^{n} x_{i}^{2}\theta_{i}^{-2}} {\sqrt{\sum_{i=1}^{n}x_{i}^{2}\theta_{i}^{-4}}} \geq \frac{\theta_{\max}\theta_{\min}}{(\theta_{\max}^{2}+\theta_{\min}^{2})/2} $

Unfortunately, I don't see any easy way to proceed from here.

Source Link
Brian Borchers
  • 3.9k
  • 1
  • 16
  • 17

Here's a restatement of the problem for those who don't want to find the paper referenced in the question.

We're interested in a symmetric and positive definitive matrix $\Sigma^{-1}$ that has been orthogonally diagonalized as

$ \Sigma=Q^{T}\mbox{diag}(\theta_{1}^{2},...,\theta_{n}^{2})Q $

where $\theta_{1}^{2} \geq \theta_{2}^{2} \geq ... \geq \theta_{n}^{2} > 0$, and we let $\theta_{\max}^{2}=\theta_{1}^{2}$ and $\theta_{\min}^{2}=\theta_{n}^{2}$.

Also, let

$\Theta^{2}=\mbox{diag}(\theta_{1}^{2},...,\theta_{n}^{2})$.

We want to show for any nonzero vector $\alpha$,

$ \frac{\alpha^{T}\Sigma^{-1}\alpha}{\sqrt{\alpha^{T}\alpha}\sqrt{\alpha^{T}\Sigma^{-2}\alpha}} \geq \frac{\theta_{\max}\theta_{\min}}{(\theta_{\max}^{2}+\theta_{\min}^{2})/2} $

Note that you can assume without loss of generality that $\alpha$ is of length 1. (just scale the length of alpha out of everything on the left hand side of the inequaility.)

Also, by using the substitution $x=Q\alpha$, you can reduce this to a problem about the digonal matrix $\Theta$, and then reduce the matrix-vector products to sums. We then want to show for all vectors $x$ of length 1,

$ \frac{\sum_{i=1}^{n} x_{i}^{2}\theta_{i}^{-2}} {\sqrt{\sum_{i=1}^{n}x_{i}^{2}\theta_{i}^{-4}}} \geq \frac{\theta_{\max}\theta_{\min}}{(\theta_{\max}^{2}+\theta_{\min}^{2})/2} $

Unfortunately, I don't see any easy way to proceed from here.