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changed inequality from strict to non-strict
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edit approved Apr 23, 2019 at 3:17
B Merlot
edit approved Apr 23, 2019 at 3:17
B Merlot
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For a positive definite diagonal matrix $A$, I want to prove that for any $x$:

$$\frac{x^T \sqrt{A} x}{\|\sqrt{A}x\|_2} > \frac{x^T A x}{\|Ax\|_2}$$$$\frac{x^T \sqrt{A} x}{\|\sqrt{A}x\|_2} \geq \frac{x^T A x}{\|Ax\|_2}$$

So far I cannot find any counterexamples, and it intuitively makes sense since the $\sqrt{\cdot}$ operator should bring the eigenvalues of $A$ closer to $1$, but I can't prove this.

EDIT: changed $>$ to $\geq$

For a positive definite diagonal matrix $A$, I want to prove that for any $x$:

$$\frac{x^T \sqrt{A} x}{\|\sqrt{A}x\|_2} > \frac{x^T A x}{\|Ax\|_2}$$

So far I cannot find any counterexamples, and it intuitively makes sense since the $\sqrt{\cdot}$ operator should bring the eigenvalues of $A$ closer to $1$, but I can't prove this.

For a positive definite diagonal matrix $A$, I want to prove that for any $x$:

$$\frac{x^T \sqrt{A} x}{\|\sqrt{A}x\|_2} \geq \frac{x^T A x}{\|Ax\|_2}$$

So far I cannot find any counterexamples, and it intuitively makes sense since the $\sqrt{\cdot}$ operator should bring the eigenvalues of $A$ closer to $1$, but I can't prove this.

EDIT: changed $>$ to $\geq$

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Reginald
Reginald
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Proving inequality for positive definite matrix

For a positive definite diagonal matrix $A$, I want to prove that for any $x$:

$$\frac{x^T \sqrt{A} x}{\|\sqrt{A}x\|_2} > \frac{x^T A x}{\|Ax\|_2}$$

So far I cannot find any counterexamples, and it intuitively makes sense since the $\sqrt{\cdot}$ operator should bring the eigenvalues of $A$ closer to $1$, but I can't prove this.