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Let $A$ be a nonsingular $n \times n$ matrix, $\|\cdot\|$ be any natural norm, and $K_{p}(A)=$ $\|A\|_{p}\left\|A^{-1}\right\|_{p} .$ Let $\lambda_{1}$ be the smallest and $\lambda_{n}$ be the largest eigenvalues of the matrix $A^{t} A$

(a) Show that if $\lambda$ is an eigenvalue of $A^{t} A$, then $0<\lambda \leq\left\|A^{t}\right\|\|A\|$.

(b) Show that $K_{2}(A)=\sqrt{\frac{\lambda_{n}}{\lambda_{1}}}$. (Hint: One should use the $\|A\|=\sqrt{\rho\left(A^{t} A\right)}=\sqrt{\rho\left(A A^{t}\right)}$ relation)

For part $(a)$, the LHS inequality seems mostly clear due to the fact that symmetric matrices have non-negative eigenvalues (although I am not sure how zero eigenvalues can not exist). However, the LHS is more obscure in terms of mathematical rigor on how properties of any norm can be applied in this inequality.

For part $(b)$, it seems minimal eigenvalue of matrix $A$ will be the largest eigenvalue of the inverse of matrix $A$, so we have to use this idea but could not understand formally why square roots are needed for $K_{2}(A)$ and how spectral radius is necessary for this part?!

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  • $\begingroup$ What is a naturally norm and what is $\| \cdot \|_p$, where can $p$ come from? $\endgroup$ Apr 8, 2021 at 10:31
  • $\begingroup$ @Ramanujan mathworld.wolfram.com/…. $\endgroup$
    – Snowball
    Apr 8, 2021 at 10:32
  • $\begingroup$ 0 can't be an eigenvalue of $A$ because $A$ is nonsingular. Now apply a similar reasoning to $A^t A$. $\endgroup$ Apr 8, 2021 at 10:32
  • $\begingroup$ @Ramanujan Appreciate your comments. Could you please summarize mathematically your statements, especially for the RHS part as well?! $\endgroup$
    – Snowball
    Apr 8, 2021 at 10:33
  • $\begingroup$ @Ramanujan You can disregard p for the sake of this problem - it is not that crucial indeed for part (a). $\endgroup$
    – Snowball
    Apr 8, 2021 at 10:40

1 Answer 1

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  1. if $\|\cdot\|$ is a natural matrix norm, then $\|\cdot\|$ is multiplicative, therefore

$$\|A^{t} A\| \le \|A^{t}\|\|A\|.$$

  1. $A$ is non-singular, hence $A^tA$ is nonsingular and symmetric, thus each eigen value $ \lambda$ is positive.

  2. There exists $x$ such that $A^tAx= \lambda x$ and $\|x\|=1.$ We get:

$$0< \lambda = \|\lambda x\|= \|A^tAx\| \le \|A^tA\| \|x\|=\|A^tA\| \le \|A^{t}\|\|A\|.$$

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  • $\begingroup$ I also became curious about the part $(b)$, and wondered why square roots are needed for the aforementioned equality regarding the $K_{2}$ value?! $\endgroup$
    – Snowball
    Apr 8, 2021 at 11:30
  • $\begingroup$ Any further suggestions about the second part?! $\endgroup$
    – Snowball
    Apr 8, 2021 at 11:56

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